Free Step-by-Step Integral Solver

1. There are multiple ways to do this integral.

   Method #1

   1. Use integration by parts:

      $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

      Let [tex]u{\left (x \right )} = 4 x^{6} - 2 x^{5} + 2 x^{3} - 10 x + 5[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

      Then [tex]\operatorname{du}{\left (x \right )} = 24 x^{5} - 10 x^{4} + 6 x^{2} - 10[/tex].

      To find [tex]v{\left (x \right )}[/tex]:

      1. Let [tex]u = 4 x[/tex].

         Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

         [tex]\int \frac{e^{u}}{4}\, du[/tex]

         1. The integral of a constant times a function is the constant times the integral of the function:

            [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

            1. The integral of the exponential function is itself.

               [tex]\int e^{u}\, du = e^{u}[/tex]

            So, the result is: [tex]\frac{e^{u}}{4}[/tex]

         Now substitute [tex]u[/tex] back in:

         [tex]\frac{e^{4 x}}{4}[/tex]

      Now evaluate the sub-integral.

   2. The integral of a constant times a function is the constant times the integral of the function:

      [tex]\int \frac{e^{4 x}}{4} \left(24 x^{5} - 10 x^{4} + 6 x^{2} - 10\right)\, dx = \frac{1}{4} \int \left(24 x^{5} - 10 x^{4} + 6 x^{2} - 10\right) e^{4 x}\, dx[/tex]

      1. Rewrite the integrand:

         [tex]\left(24 x^{5} - 10 x^{4} + 6 x^{2} - 10\right) e^{4 x} = 24 x^{5} e^{4 x} - 10 x^{4} e^{4 x} + 6 x^{2} e^{4 x} - 10 e^{4 x}[/tex]

      2. Integrate term-by-term:

         1. The integral of a constant times a function is the constant times the integral of the function:

            [tex]\int 24 x^{5} e^{4 x}\, dx = 24 \int x^{5} e^{4 x}\, dx[/tex]

            1. Use integration by parts:

               $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

               Let [tex]u{\left (x \right )} = x^{5}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

               Then [tex]\operatorname{du}{\left (x \right )} = 5 x^{4}[/tex].

               To find [tex]v{\left (x \right )}[/tex]:

               1. Let [tex]u = 4 x[/tex].

                  Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

                  [tex]\int \frac{e^{u}}{4}\, du[/tex]

                  1. The integral of a constant times a function is the constant times the integral of the function:

                     [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                     1. The integral of the exponential function is itself.

                        [tex]\int e^{u}\, du = e^{u}[/tex]

                     So, the result is: [tex]\frac{e^{u}}{4}[/tex]

                  Now substitute [tex]u[/tex] back in:

                  [tex]\frac{e^{4 x}}{4}[/tex]

               Now evaluate the sub-integral.

            2. Use integration by parts:

               $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

               Let [tex]u{\left (x \right )} = \frac{5 x^{4}}{4}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

               Then [tex]\operatorname{du}{\left (x \right )} = 5 x^{3}[/tex].

               To find [tex]v{\left (x \right )}[/tex]:

               1. Let [tex]u = 4 x[/tex].

                  Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

                  [tex]\int \frac{e^{u}}{4}\, du[/tex]

                  1. The integral of a constant times a function is the constant times the integral of the function:

                     [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                     1. The integral of the exponential function is itself.

                        [tex]\int e^{u}\, du = e^{u}[/tex]

                     So, the result is: [tex]\frac{e^{u}}{4}[/tex]

                  Now substitute [tex]u[/tex] back in:

                  [tex]\frac{e^{4 x}}{4}[/tex]

               Now evaluate the sub-integral.

            3. Use integration by parts:

               $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

               Let [tex]u{\left (x \right )} = \frac{5 x^{3}}{4}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

               Then [tex]\operatorname{du}{\left (x \right )} = \frac{15 x^{2}}{4}[/tex].

               To find [tex]v{\left (x \right )}[/tex]:

               1. Let [tex]u = 4 x[/tex].

                  Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

                  [tex]\int \frac{e^{u}}{4}\, du[/tex]

                  1. The integral of a constant times a function is the constant times the integral of the function:

                     [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                     1. The integral of the exponential function is itself.

                        [tex]\int e^{u}\, du = e^{u}[/tex]

                     So, the result is: [tex]\frac{e^{u}}{4}[/tex]

                  Now substitute [tex]u[/tex] back in:

                  [tex]\frac{e^{4 x}}{4}[/tex]

               Now evaluate the sub-integral.

            4. Use integration by parts:

               $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

               Let [tex]u{\left (x \right )} = \frac{15 x^{2}}{16}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

               Then [tex]\operatorname{du}{\left (x \right )} = \frac{15 x}{8}[/tex].

               To find [tex]v{\left (x \right )}[/tex]:

               1. Let [tex]u = 4 x[/tex].

                  Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

                  [tex]\int \frac{e^{u}}{4}\, du[/tex]

                  1. The integral of a constant times a function is the constant times the integral of the function:

                     [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                     1. The integral of the exponential function is itself.

                        [tex]\int e^{u}\, du = e^{u}[/tex]

                     So, the result is: [tex]\frac{e^{u}}{4}[/tex]

                  Now substitute [tex]u[/tex] back in:

                  [tex]\frac{e^{4 x}}{4}[/tex]

               Now evaluate the sub-integral.

            5. Use integration by parts:

               $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

               Let [tex]u{\left (x \right )} = \frac{15 x}{32}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

               Then [tex]\operatorname{du}{\left (x \right )} = \frac{15}{32}[/tex].

               To find [tex]v{\left (x \right )}[/tex]:

               1. Let [tex]u = 4 x[/tex].

                  Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

                  [tex]\int \frac{e^{u}}{4}\, du[/tex]

                  1. The integral of a constant times a function is the constant times the integral of the function:

                     [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                     1. The integral of the exponential function is itself.

                        [tex]\int e^{u}\, du = e^{u}[/tex]

                     So, the result is: [tex]\frac{e^{u}}{4}[/tex]

                  Now substitute [tex]u[/tex] back in:

                  [tex]\frac{e^{4 x}}{4}[/tex]

               Now evaluate the sub-integral.

            6. The integral of a constant times a function is the constant times the integral of the function:

               [tex]\int \frac{15}{128} e^{4 x}\, dx = \frac{15}{128} \int e^{4 x}\, dx[/tex]

               1. Let [tex]u = 4 x[/tex].

                  Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

                  [tex]\int \frac{e^{u}}{4}\, du[/tex]

                  1. The integral of a constant times a function is the constant times the integral of the function:

                     [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                     1. The integral of the exponential function is itself.

                        [tex]\int e^{u}\, du = e^{u}[/tex]

                     So, the result is: [tex]\frac{e^{u}}{4}[/tex]

                  Now substitute [tex]u[/tex] back in:

                  [tex]\frac{e^{4 x}}{4}[/tex]

               So, the result is: [tex]\frac{15}{512} e^{4 x}[/tex]

            So, the result is: [tex]6 x^{5} e^{4 x} - \frac{15 x^{4}}{2} e^{4 x} + \frac{15 x^{3}}{2} e^{4 x} - \frac{45 x^{2}}{8} e^{4 x} + \frac{45 x}{16} e^{4 x} - \frac{45}{64} e^{4 x}[/tex]

         1. The integral of a constant times a function is the constant times the integral of the function:

            [tex]\int - 10 x^{4} e^{4 x}\, dx = - 10 \int x^{4} e^{4 x}\, dx[/tex]

            1. Use integration by parts:

               $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

               Let [tex]u{\left (x \right )} = x^{4}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

               Then [tex]\operatorname{du}{\left (x \right )} = 4 x^{3}[/tex].

               To find [tex]v{\left (x \right )}[/tex]:

               1. Let [tex]u = 4 x[/tex].

                  Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

                  [tex]\int \frac{e^{u}}{4}\, du[/tex]

                  1. The integral of a constant times a function is the constant times the integral of the function:

                     [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                     1. The integral of the exponential function is itself.

                        [tex]\int e^{u}\, du = e^{u}[/tex]

                     So, the result is: [tex]\frac{e^{u}}{4}[/tex]

                  Now substitute [tex]u[/tex] back in:

                  [tex]\frac{e^{4 x}}{4}[/tex]

               Now evaluate the sub-integral.

            2. Use integration by parts:

               $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

               Let [tex]u{\left (x \right )} = x^{3}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

               Then [tex]\operatorname{du}{\left (x \right )} = 3 x^{2}[/tex].

               To find [tex]v{\left (x \right )}[/tex]:

               1. Let [tex]u = 4 x[/tex].

                  Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

                  [tex]\int \frac{e^{u}}{4}\, du[/tex]

                  1. The integral of a constant times a function is the constant times the integral of the function:

                     [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                     1. The integral of the exponential function is itself.

                        [tex]\int e^{u}\, du = e^{u}[/tex]

                     So, the result is: [tex]\frac{e^{u}}{4}[/tex]

                  Now substitute [tex]u[/tex] back in:

                  [tex]\frac{e^{4 x}}{4}[/tex]

               Now evaluate the sub-integral.

            3. Use integration by parts:

               $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

               Let [tex]u{\left (x \right )} = \frac{3 x^{2}}{4}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

               Then [tex]\operatorname{du}{\left (x \right )} = \frac{3 x}{2}[/tex].

               To find [tex]v{\left (x \right )}[/tex]:

               1. Let [tex]u = 4 x[/tex].

                  Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

                  [tex]\int \frac{e^{u}}{4}\, du[/tex]

                  1. The integral of a constant times a function is the constant times the integral of the function:

                     [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                     1. The integral of the exponential function is itself.

                        [tex]\int e^{u}\, du = e^{u}[/tex]

                     So, the result is: [tex]\frac{e^{u}}{4}[/tex]

                  Now substitute [tex]u[/tex] back in:

                  [tex]\frac{e^{4 x}}{4}[/tex]

               Now evaluate the sub-integral.

            4. Use integration by parts:

               $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

               Let [tex]u{\left (x \right )} = \frac{3 x}{8}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

               Then [tex]\operatorname{du}{\left (x \right )} = \frac{3}{8}[/tex].

               To find [tex]v{\left (x \right )}[/tex]:

               1. Let [tex]u = 4 x[/tex].

                  Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

                  [tex]\int \frac{e^{u}}{4}\, du[/tex]

                  1. The integral of a constant times a function is the constant times the integral of the function:

                     [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                     1. The integral of the exponential function is itself.

                        [tex]\int e^{u}\, du = e^{u}[/tex]

                     So, the result is: [tex]\frac{e^{u}}{4}[/tex]

                  Now substitute [tex]u[/tex] back in:

                  [tex]\frac{e^{4 x}}{4}[/tex]

               Now evaluate the sub-integral.

            5. The integral of a constant times a function is the constant times the integral of the function:

               [tex]\int \frac{3}{32} e^{4 x}\, dx = \frac{3}{32} \int e^{4 x}\, dx[/tex]

               1. Let [tex]u = 4 x[/tex].

                  Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

                  [tex]\int \frac{e^{u}}{4}\, du[/tex]

                  1. The integral of a constant times a function is the constant times the integral of the function:

                     [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                     1. The integral of the exponential function is itself.

                        [tex]\int e^{u}\, du = e^{u}[/tex]

                     So, the result is: [tex]\frac{e^{u}}{4}[/tex]

                  Now substitute [tex]u[/tex] back in:

                  [tex]\frac{e^{4 x}}{4}[/tex]

               So, the result is: [tex]\frac{3}{128} e^{4 x}[/tex]

            So, the result is: [tex]- \frac{5 x^{4}}{2} e^{4 x} + \frac{5 x^{3}}{2} e^{4 x} - \frac{15 x^{2}}{8} e^{4 x} + \frac{15 x}{16} e^{4 x} - \frac{15}{64} e^{4 x}[/tex]

         1. The integral of a constant times a function is the constant times the integral of the function:

            [tex]\int 6 x^{2} e^{4 x}\, dx = 6 \int x^{2} e^{4 x}\, dx[/tex]

            1. Use integration by parts:

               $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

               Let [tex]u{\left (x \right )} = x^{2}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

               Then [tex]\operatorname{du}{\left (x \right )} = 2 x[/tex].

               To find [tex]v{\left (x \right )}[/tex]:

               1. Let [tex]u = 4 x[/tex].

                  Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

                  [tex]\int \frac{e^{u}}{4}\, du[/tex]

                  1. The integral of a constant times a function is the constant times the integral of the function:

                     [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                     1. The integral of the exponential function is itself.

                        [tex]\int e^{u}\, du = e^{u}[/tex]

                     So, the result is: [tex]\frac{e^{u}}{4}[/tex]

                  Now substitute [tex]u[/tex] back in:

                  [tex]\frac{e^{4 x}}{4}[/tex]

               Now evaluate the sub-integral.

            2. Use integration by parts:

               $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

               Let [tex]u{\left (x \right )} = \frac{x}{2}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

               Then [tex]\operatorname{du}{\left (x \right )} = \frac{1}{2}[/tex].

               To find [tex]v{\left (x \right )}[/tex]:

               1. Let [tex]u = 4 x[/tex].

                  Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

                  [tex]\int \frac{e^{u}}{4}\, du[/tex]

                  1. The integral of a constant times a function is the constant times the integral of the function:

                     [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                     1. The integral of the exponential function is itself.

                        [tex]\int e^{u}\, du = e^{u}[/tex]

                     So, the result is: [tex]\frac{e^{u}}{4}[/tex]

                  Now substitute [tex]u[/tex] back in:

                  [tex]\frac{e^{4 x}}{4}[/tex]

               Now evaluate the sub-integral.

            3. The integral of a constant times a function is the constant times the integral of the function:

               [tex]\int \frac{e^{4 x}}{8}\, dx = \frac{1}{8} \int e^{4 x}\, dx[/tex]

               1. Let [tex]u = 4 x[/tex].

                  Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

                  [tex]\int \frac{e^{u}}{4}\, du[/tex]

                  1. The integral of a constant times a function is the constant times the integral of the function:

                     [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                     1. The integral of the exponential function is itself.

                        [tex]\int e^{u}\, du = e^{u}[/tex]

                     So, the result is: [tex]\frac{e^{u}}{4}[/tex]

                  Now substitute [tex]u[/tex] back in:

                  [tex]\frac{e^{4 x}}{4}[/tex]

               So, the result is: [tex]\frac{e^{4 x}}{32}[/tex]

            So, the result is: [tex]\frac{3 x^{2}}{2} e^{4 x} - \frac{3 x}{4} e^{4 x} + \frac{3}{16} e^{4 x}[/tex]

         1. The integral of a constant times a function is the constant times the integral of the function:

            [tex]\int - 10 e^{4 x}\, dx = - 10 \int e^{4 x}\, dx[/tex]

            1. Let [tex]u = 4 x[/tex].

               Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

               [tex]\int \frac{e^{u}}{4}\, du[/tex]

               1. The integral of a constant times a function is the constant times the integral of the function:

                  [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                  1. The integral of the exponential function is itself.

                     [tex]\int e^{u}\, du = e^{u}[/tex]

                  So, the result is: [tex]\frac{e^{u}}{4}[/tex]

               Now substitute [tex]u[/tex] back in:

               [tex]\frac{e^{4 x}}{4}[/tex]

            So, the result is: [tex]- \frac{5}{2} e^{4 x}[/tex]

         The result is: [tex]6 x^{5} e^{4 x} - 10 x^{4} e^{4 x} + 10 x^{3} e^{4 x} - 6 x^{2} e^{4 x} + 3 x e^{4 x} - \frac{13}{4} e^{4 x}[/tex]

      So, the result is: [tex]\frac{3 x^{5}}{2} e^{4 x} - \frac{5 x^{4}}{2} e^{4 x} + \frac{5 x^{3}}{2} e^{4 x} - \frac{3 x^{2}}{2} e^{4 x} + \frac{3 x}{4} e^{4 x} - \frac{13}{16} e^{4 x}[/tex]

   Method #2

   1. Rewrite the integrand:

      [tex]\left(4 x^{6} - 2 x^{5} + 2 x^{3} - 10 x + 5\right) e^{4 x} = 4 x^{6} e^{4 x} - 2 x^{5} e^{4 x} + 2 x^{3} e^{4 x} - 10 x e^{4 x} + 5 e^{4 x}[/tex]

   2. Integrate term-by-term:

      1. The integral of a constant times a function is the constant times the integral of the function:

         [tex]\int 4 x^{6} e^{4 x}\, dx = 4 \int x^{6} e^{4 x}\, dx[/tex]

         1. Use integration by parts:

            $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

            Let [tex]u{\left (x \right )} = x^{6}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

            Then [tex]\operatorname{du}{\left (x \right )} = 6 x^{5}[/tex].

            To find [tex]v{\left (x \right )}[/tex]:

            1. Let [tex]u = 4 x[/tex].

               Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

               [tex]\int \frac{e^{u}}{4}\, du[/tex]

               1. The integral of a constant times a function is the constant times the integral of the function:

                  [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                  1. The integral of the exponential function is itself.

                     [tex]\int e^{u}\, du = e^{u}[/tex]

                  So, the result is: [tex]\frac{e^{u}}{4}[/tex]

               Now substitute [tex]u[/tex] back in:

               [tex]\frac{e^{4 x}}{4}[/tex]

            Now evaluate the sub-integral.

         2. Use integration by parts:

            $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

            Let [tex]u{\left (x \right )} = \frac{3 x^{5}}{2}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

            Then [tex]\operatorname{du}{\left (x \right )} = \frac{15 x^{4}}{2}[/tex].

            To find [tex]v{\left (x \right )}[/tex]:

            1. Let [tex]u = 4 x[/tex].

               Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

               [tex]\int \frac{e^{u}}{4}\, du[/tex]

               1. The integral of a constant times a function is the constant times the integral of the function:

                  [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                  1. The integral of the exponential function is itself.

                     [tex]\int e^{u}\, du = e^{u}[/tex]

                  So, the result is: [tex]\frac{e^{u}}{4}[/tex]

               Now substitute [tex]u[/tex] back in:

               [tex]\frac{e^{4 x}}{4}[/tex]

            Now evaluate the sub-integral.

         3. Use integration by parts:

            $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

            Let [tex]u{\left (x \right )} = \frac{15 x^{4}}{8}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

            Then [tex]\operatorname{du}{\left (x \right )} = \frac{15 x^{3}}{2}[/tex].

            To find [tex]v{\left (x \right )}[/tex]:

            1. Let [tex]u = 4 x[/tex].

               Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

               [tex]\int \frac{e^{u}}{4}\, du[/tex]

               1. The integral of a constant times a function is the constant times the integral of the function:

                  [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                  1. The integral of the exponential function is itself.

                     [tex]\int e^{u}\, du = e^{u}[/tex]

                  So, the result is: [tex]\frac{e^{u}}{4}[/tex]

               Now substitute [tex]u[/tex] back in:

               [tex]\frac{e^{4 x}}{4}[/tex]

            Now evaluate the sub-integral.

         4. Use integration by parts:

            $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

            Let [tex]u{\left (x \right )} = \frac{15 x^{3}}{8}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

            Then [tex]\operatorname{du}{\left (x \right )} = \frac{45 x^{2}}{8}[/tex].

            To find [tex]v{\left (x \right )}[/tex]:

            1. Let [tex]u = 4 x[/tex].

               Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

               [tex]\int \frac{e^{u}}{4}\, du[/tex]

               1. The integral of a constant times a function is the constant times the integral of the function:

                  [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                  1. The integral of the exponential function is itself.

                     [tex]\int e^{u}\, du = e^{u}[/tex]

                  So, the result is: [tex]\frac{e^{u}}{4}[/tex]

               Now substitute [tex]u[/tex] back in:

               [tex]\frac{e^{4 x}}{4}[/tex]

            Now evaluate the sub-integral.

         5. Use integration by parts:

            $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

            Let [tex]u{\left (x \right )} = \frac{45 x^{2}}{32}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

            Then [tex]\operatorname{du}{\left (x \right )} = \frac{45 x}{16}[/tex].

            To find [tex]v{\left (x \right )}[/tex]:

            1. Let [tex]u = 4 x[/tex].

               Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

               [tex]\int \frac{e^{u}}{4}\, du[/tex]

               1. The integral of a constant times a function is the constant times the integral of the function:

                  [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                  1. The integral of the exponential function is itself.

                     [tex]\int e^{u}\, du = e^{u}[/tex]

                  So, the result is: [tex]\frac{e^{u}}{4}[/tex]

               Now substitute [tex]u[/tex] back in:

               [tex]\frac{e^{4 x}}{4}[/tex]

            Now evaluate the sub-integral.

         6. Use integration by parts:

            $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

            Let [tex]u{\left (x \right )} = \frac{45 x}{64}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

            Then [tex]\operatorname{du}{\left (x \right )} = \frac{45}{64}[/tex].

            To find [tex]v{\left (x \right )}[/tex]:

            1. Let [tex]u = 4 x[/tex].

               Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

               [tex]\int \frac{e^{u}}{4}\, du[/tex]

               1. The integral of a constant times a function is the constant times the integral of the function:

                  [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                  1. The integral of the exponential function is itself.

                     [tex]\int e^{u}\, du = e^{u}[/tex]

                  So, the result is: [tex]\frac{e^{u}}{4}[/tex]

               Now substitute [tex]u[/tex] back in:

               [tex]\frac{e^{4 x}}{4}[/tex]

            Now evaluate the sub-integral.

         7. The integral of a constant times a function is the constant times the integral of the function:

            [tex]\int \frac{45}{256} e^{4 x}\, dx = \frac{45}{256} \int e^{4 x}\, dx[/tex]

            1. Let [tex]u = 4 x[/tex].

               Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

               [tex]\int \frac{e^{u}}{4}\, du[/tex]

               1. The integral of a constant times a function is the constant times the integral of the function:

                  [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                  1. The integral of the exponential function is itself.

                     [tex]\int e^{u}\, du = e^{u}[/tex]

                  So, the result is: [tex]\frac{e^{u}}{4}[/tex]

               Now substitute [tex]u[/tex] back in:

               [tex]\frac{e^{4 x}}{4}[/tex]

            So, the result is: [tex]\frac{45}{1024} e^{4 x}[/tex]

         So, the result is: [tex]x^{6} e^{4 x} - \frac{3 x^{5}}{2} e^{4 x} + \frac{15 x^{4}}{8} e^{4 x} - \frac{15 x^{3}}{8} e^{4 x} + \frac{45 x^{2}}{32} e^{4 x} - \frac{45 x}{64} e^{4 x} + \frac{45}{256} e^{4 x}[/tex]

      1. The integral of a constant times a function is the constant times the integral of the function:

         [tex]\int - 2 x^{5} e^{4 x}\, dx = - 2 \int x^{5} e^{4 x}\, dx[/tex]

         1. Use integration by parts:

            $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

            Let [tex]u{\left (x \right )} = x^{5}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

            Then [tex]\operatorname{du}{\left (x \right )} = 5 x^{4}[/tex].

            To find [tex]v{\left (x \right )}[/tex]:

            1. Let [tex]u = 4 x[/tex].

               Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

               [tex]\int \frac{e^{u}}{4}\, du[/tex]

               1. The integral of a constant times a function is the constant times the integral of the function:

                  [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                  1. The integral of the exponential function is itself.

                     [tex]\int e^{u}\, du = e^{u}[/tex]

                  So, the result is: [tex]\frac{e^{u}}{4}[/tex]

               Now substitute [tex]u[/tex] back in:

               [tex]\frac{e^{4 x}}{4}[/tex]

            Now evaluate the sub-integral.

         2. Use integration by parts:

            $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

            Let [tex]u{\left (x \right )} = \frac{5 x^{4}}{4}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

            Then [tex]\operatorname{du}{\left (x \right )} = 5 x^{3}[/tex].

            To find [tex]v{\left (x \right )}[/tex]:

            1. Let [tex]u = 4 x[/tex].

               Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

               [tex]\int \frac{e^{u}}{4}\, du[/tex]

               1. The integral of a constant times a function is the constant times the integral of the function:

                  [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                  1. The integral of the exponential function is itself.

                     [tex]\int e^{u}\, du = e^{u}[/tex]

                  So, the result is: [tex]\frac{e^{u}}{4}[/tex]

               Now substitute [tex]u[/tex] back in:

               [tex]\frac{e^{4 x}}{4}[/tex]

            Now evaluate the sub-integral.

         3. Use integration by parts:

            $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

            Let [tex]u{\left (x \right )} = \frac{5 x^{3}}{4}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

            Then [tex]\operatorname{du}{\left (x \right )} = \frac{15 x^{2}}{4}[/tex].

            To find [tex]v{\left (x \right )}[/tex]:

            1. Let [tex]u = 4 x[/tex].

               Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

               [tex]\int \frac{e^{u}}{4}\, du[/tex]

               1. The integral of a constant times a function is the constant times the integral of the function:

                  [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                  1. The integral of the exponential function is itself.

                     [tex]\int e^{u}\, du = e^{u}[/tex]

                  So, the result is: [tex]\frac{e^{u}}{4}[/tex]

               Now substitute [tex]u[/tex] back in:

               [tex]\frac{e^{4 x}}{4}[/tex]

            Now evaluate the sub-integral.

         4. Use integration by parts:

            $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

            Let [tex]u{\left (x \right )} = \frac{15 x^{2}}{16}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

            Then [tex]\operatorname{du}{\left (x \right )} = \frac{15 x}{8}[/tex].

            To find [tex]v{\left (x \right )}[/tex]:

            1. Let [tex]u = 4 x[/tex].

               Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

               [tex]\int \frac{e^{u}}{4}\, du[/tex]

               1. The integral of a constant times a function is the constant times the integral of the function:

                  [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                  1. The integral of the exponential function is itself.

                     [tex]\int e^{u}\, du = e^{u}[/tex]

                  So, the result is: [tex]\frac{e^{u}}{4}[/tex]

               Now substitute [tex]u[/tex] back in:

               [tex]\frac{e^{4 x}}{4}[/tex]

            Now evaluate the sub-integral.

         5. Use integration by parts:

            $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

            Let [tex]u{\left (x \right )} = \frac{15 x}{32}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

            Then [tex]\operatorname{du}{\left (x \right )} = \frac{15}{32}[/tex].

            To find [tex]v{\left (x \right )}[/tex]:

            1. Let [tex]u = 4 x[/tex].

               Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

               [tex]\int \frac{e^{u}}{4}\, du[/tex]

               1. The integral of a constant times a function is the constant times the integral of the function:

                  [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                  1. The integral of the exponential function is itself.

                     [tex]\int e^{u}\, du = e^{u}[/tex]

                  So, the result is: [tex]\frac{e^{u}}{4}[/tex]

               Now substitute [tex]u[/tex] back in:

               [tex]\frac{e^{4 x}}{4}[/tex]

            Now evaluate the sub-integral.

         6. The integral of a constant times a function is the constant times the integral of the function:

            [tex]\int \frac{15}{128} e^{4 x}\, dx = \frac{15}{128} \int e^{4 x}\, dx[/tex]

            1. Let [tex]u = 4 x[/tex].

               Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

               [tex]\int \frac{e^{u}}{4}\, du[/tex]

               1. The integral of a constant times a function is the constant times the integral of the function:

                  [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                  1. The integral of the exponential function is itself.

                     [tex]\int e^{u}\, du = e^{u}[/tex]

                  So, the result is: [tex]\frac{e^{u}}{4}[/tex]

               Now substitute [tex]u[/tex] back in:

               [tex]\frac{e^{4 x}}{4}[/tex]

            So, the result is: [tex]\frac{15}{512} e^{4 x}[/tex]

         So, the result is: [tex]- \frac{x^{5}}{2} e^{4 x} + \frac{5 x^{4}}{8} e^{4 x} - \frac{5 x^{3}}{8} e^{4 x} + \frac{15 x^{2}}{32} e^{4 x} - \frac{15 x}{64} e^{4 x} + \frac{15}{256} e^{4 x}[/tex]

      1. The integral of a constant times a function is the constant times the integral of the function:

         [tex]\int 2 x^{3} e^{4 x}\, dx = 2 \int x^{3} e^{4 x}\, dx[/tex]

         1. Use integration by parts:

            $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

            Let [tex]u{\left (x \right )} = x^{3}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

            Then [tex]\operatorname{du}{\left (x \right )} = 3 x^{2}[/tex].

            To find [tex]v{\left (x \right )}[/tex]:

            1. Let [tex]u = 4 x[/tex].

               Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

               [tex]\int \frac{e^{u}}{4}\, du[/tex]

               1. The integral of a constant times a function is the constant times the integral of the function:

                  [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                  1. The integral of the exponential function is itself.

                     [tex]\int e^{u}\, du = e^{u}[/tex]

                  So, the result is: [tex]\frac{e^{u}}{4}[/tex]

               Now substitute [tex]u[/tex] back in:

               [tex]\frac{e^{4 x}}{4}[/tex]

            Now evaluate the sub-integral.

         2. Use integration by parts:

            $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

            Let [tex]u{\left (x \right )} = \frac{3 x^{2}}{4}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

            Then [tex]\operatorname{du}{\left (x \right )} = \frac{3 x}{2}[/tex].

            To find [tex]v{\left (x \right )}[/tex]:

            1. Let [tex]u = 4 x[/tex].

               Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

               [tex]\int \frac{e^{u}}{4}\, du[/tex]

               1. The integral of a constant times a function is the constant times the integral of the function:

                  [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                  1. The integral of the exponential function is itself.

                     [tex]\int e^{u}\, du = e^{u}[/tex]

                  So, the result is: [tex]\frac{e^{u}}{4}[/tex]

               Now substitute [tex]u[/tex] back in:

               [tex]\frac{e^{4 x}}{4}[/tex]

            Now evaluate the sub-integral.

         3. Use integration by parts:

            $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

            Let [tex]u{\left (x \right )} = \frac{3 x}{8}[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

            Then [tex]\operatorname{du}{\left (x \right )} = \frac{3}{8}[/tex].

            To find [tex]v{\left (x \right )}[/tex]:

            1. Let [tex]u = 4 x[/tex].

               Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

               [tex]\int \frac{e^{u}}{4}\, du[/tex]

               1. The integral of a constant times a function is the constant times the integral of the function:

                  [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                  1. The integral of the exponential function is itself.

                     [tex]\int e^{u}\, du = e^{u}[/tex]

                  So, the result is: [tex]\frac{e^{u}}{4}[/tex]

               Now substitute [tex]u[/tex] back in:

               [tex]\frac{e^{4 x}}{4}[/tex]

            Now evaluate the sub-integral.

         4. The integral of a constant times a function is the constant times the integral of the function:

            [tex]\int \frac{3}{32} e^{4 x}\, dx = \frac{3}{32} \int e^{4 x}\, dx[/tex]

            1. Let [tex]u = 4 x[/tex].

               Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

               [tex]\int \frac{e^{u}}{4}\, du[/tex]

               1. The integral of a constant times a function is the constant times the integral of the function:

                  [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                  1. The integral of the exponential function is itself.

                     [tex]\int e^{u}\, du = e^{u}[/tex]

                  So, the result is: [tex]\frac{e^{u}}{4}[/tex]

               Now substitute [tex]u[/tex] back in:

               [tex]\frac{e^{4 x}}{4}[/tex]

            So, the result is: [tex]\frac{3}{128} e^{4 x}[/tex]

         So, the result is: [tex]\frac{x^{3}}{2} e^{4 x} - \frac{3 x^{2}}{8} e^{4 x} + \frac{3 x}{16} e^{4 x} - \frac{3}{64} e^{4 x}[/tex]

      1. The integral of a constant times a function is the constant times the integral of the function:

         [tex]\int - 10 x e^{4 x}\, dx = - 10 \int x e^{4 x}\, dx[/tex]

         1. Use integration by parts:

            $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

            Let [tex]u{\left (x \right )} = x[/tex] and let [tex]\operatorname{dv}{\left (x \right )} = e^{4 x}[/tex].

            Then [tex]\operatorname{du}{\left (x \right )} = 1[/tex].

            To find [tex]v{\left (x \right )}[/tex]:

            1. Let [tex]u = 4 x[/tex].

               Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

               [tex]\int \frac{e^{u}}{4}\, du[/tex]

               1. The integral of a constant times a function is the constant times the integral of the function:

                  [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                  1. The integral of the exponential function is itself.

                     [tex]\int e^{u}\, du = e^{u}[/tex]

                  So, the result is: [tex]\frac{e^{u}}{4}[/tex]

               Now substitute [tex]u[/tex] back in:

               [tex]\frac{e^{4 x}}{4}[/tex]

            Now evaluate the sub-integral.

         2. The integral of a constant times a function is the constant times the integral of the function:

            [tex]\int \frac{e^{4 x}}{4}\, dx = \frac{1}{4} \int e^{4 x}\, dx[/tex]

            1. Let [tex]u = 4 x[/tex].

               Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

               [tex]\int \frac{e^{u}}{4}\, du[/tex]

               1. The integral of a constant times a function is the constant times the integral of the function:

                  [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

                  1. The integral of the exponential function is itself.

                     [tex]\int e^{u}\, du = e^{u}[/tex]

                  So, the result is: [tex]\frac{e^{u}}{4}[/tex]

               Now substitute [tex]u[/tex] back in:

               [tex]\frac{e^{4 x}}{4}[/tex]

            So, the result is: [tex]\frac{e^{4 x}}{16}[/tex]

         So, the result is: [tex]- \frac{5 x}{2} e^{4 x} + \frac{5}{8} e^{4 x}[/tex]

      1. The integral of a constant times a function is the constant times the integral of the function:

         [tex]\int 5 e^{4 x}\, dx = 5 \int e^{4 x}\, dx[/tex]

         1. Let [tex]u = 4 x[/tex].

            Then let [tex]du = 4 dx[/tex] and substitute [tex]\frac{du}{4}[/tex]:

            [tex]\int \frac{e^{u}}{4}\, du[/tex]

            1. The integral of a constant times a function is the constant times the integral of the function:

               [tex]\int e^{u}\, du = \frac{1}{4} \int e^{u}\, du[/tex]

               1. The integral of the exponential function is itself.

                  [tex]\int e^{u}\, du = e^{u}[/tex]

               So, the result is: [tex]\frac{e^{u}}{4}[/tex]

            Now substitute [tex]u[/tex] back in:

            [tex]\frac{e^{4 x}}{4}[/tex]

         So, the result is: [tex]\frac{5}{4} e^{4 x}[/tex]

      The result is: [tex]x^{6} e^{4 x} - 2 x^{5} e^{4 x} + \frac{5 x^{4}}{2} e^{4 x} - 2 x^{3} e^{4 x} + \frac{3 x^{2}}{2} e^{4 x} - \frac{13 x}{4} e^{4 x} + \frac{33}{16} e^{4 x}[/tex]

2. Now simplify:

   [tex]\frac{e^{4 x}}{16} \left(16 x^{6} - 32 x^{5} + 40 x^{4} - 32 x^{3} + 24 x^{2} - 52 x + 33\right)[/tex]

3. Add the constant of integration:

   [tex]\frac{e^{4 x}}{16} \left(16 x^{6} - 32 x^{5} + 40 x^{4} - 32 x^{3} + 24 x^{2} - 52 x + 33\right)+ \mathrm{constant}[/tex]

The answer is:

[tex]\frac{e^{4 x}}{16} \left(16 x^{6} - 32 x^{5} + 40 x^{4} - 32 x^{3} + 24 x^{2} - 52 x + 33\right)+ \mathrm{constant}[/tex]