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Integral
dx
[tex]\int x^{2} + 36\, dx = \frac{x^{3}}{3} + 36 x + \mathrm{const}[/tex]

Integral Steps:

  1. Integrate term-by-term:

    1. The integral of [tex]x^{n}[/tex] is [tex]\frac{x^{n + 1}}{n + 1}[/tex] when [tex]n \neq -1[/tex]:

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

    1. The integral of a constant is the constant times the variable of integration:

      [tex]\int 36\, dx = 36 x[/tex]

    The result is: [tex]\frac{x^{3}}{3} + 36 x[/tex]

  2. Now simplify:

    [tex]\frac{x}{3} \left(x^{2} + 108\right)[/tex]

  3. Add the constant of integration:

    [tex]\frac{x}{3} \left(x^{2} + 108\right)+ \mathrm{constant}[/tex]


The answer is:

[tex]\frac{x}{3} \left(x^{2} + 108\right)+ \mathrm{constant}[/tex]

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Use latex commands:
* is multiplication
oo is $\infty$
pi is $\pi$
x^2 is x2
sqrt(x) is $\sqrt{x}$
sqrt[3](x) is $\sqrt[3]{x}$
(a+b)/(c+d) is $\frac{a+b}{c+d}$

Common Integrals

$\int 0dx = \text{const}$

$\int\ dx = x + \text{const}$

$\int kdx = kx + \text{const}$

$\int x^n\ dx = \frac{1}{n+1}x^{n+1} + \text{const}$ here n≠-1

$\int \frac{1}{x}\ dx = \int x^{-1}\ dx = \ln|x| + \text{const}$

$\int x^{-n}\ dx = \frac{1}{-n+1}x^{-n+1} + \text{const}$

$\int \frac{1}{ax+b}\ dx = \frac{1}{a}\ln|ax+b| + \text{const}$

$\int e^x\ dx = e^x + \text{const}$

$\int a^x\ dx = \frac{a^x}{\\ln a} + \text{const}$

$\int \sin(x)\ dx = -\cos(x) + \text{const}$

$\int \cos(x)\ dx = \sin(x) + \text{const}$

$\int \tan(x)\ dx = \ln|sec(x)| + \text{const}$

$\int \cot(x)\ dx = \ln|\sin(x)| + \text{const}$

$\int \frac{1}{\sqrt{1-x^2}} \ dx = \arcsin(x) + \text{const}$

$\int -\frac{1}{\sqrt{1-x^2}} \ dx = \arccos(x) + \text{const}$

$\int \frac{1}{1+ x^2}\ dx = \arctan(x) + \text{const}$

$\int -\frac{1}{1+x^2}\ dx = \text{arccot}(x) + \text{const}$

Integration by Parts

$\int u\ dv = uv - \int v\ du$

$\int\limits_{a}^{b} u\ dv = uv |_a^b - \int v\ du$

Trigonometric Substitutions

$\sqrt{a^2 - b^2x^2}$ $\Rightarrow x=\frac{a}{b}\sin\theta$ and $\cos^2\theta = 1 - \sin^2\theta$

$\sqrt{a^2 + b^2x^2}$ $\Rightarrow x=\frac{a}{b}\tan\theta$ and $\sec^2\theta = 1 + \tan^2\theta$

$\sqrt{b^2x^2 - a^2}$ $\Rightarrow x=\frac{a}{b}\sec\theta$ and $\tan^2\theta = \sec^2\theta - 1$

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