Indefinite Integrals

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Definition of an Indefinite Integral

If $\frac{dy}{dx}=f(x)$ then y is the function whose derivative is $f(x)$ and is called anti - derivative of $f(x)$ or the indefinite integral of $f(x)$, denoted by $\int f(x)\ dx$. Similarly if $y = \int f(u) \ du$, then $\frac{dy}{du}=f(u)$. Since the derivative of a constant is zero, all indefinite integrals differ by an arbitrary constant.

The process of finding an integral is called integration.

General Rules of Integration

In the following $u, v, w$ are functions of $x; a, b, p, q, n$ any constants, restricted if indicated; $e = 2.71828...$ is the natural base of logarithms; $\ln u$ denotes the natural logarithm of $u$ where it is assumed that $u > 0$ [in general, to extend formulas to cases where $u > 0$ as well, replace $\ln u$ by $\ln |u|$]; all angles are in radians; all constants of integration are omitted but implied.

$\int a \ dx = ax$

$\int a f(x) \ dx = a \int f(x) \ dx$

$\int (u \pm v \pm w \pm \cdot \cdot \cdot ) \ dx = \int u \ dx \pm \int v \ dx \pm \int w \ dx \pm \cdot \cdot \cdot$

$\int u \ dv = u.v - \int v \ du \ [integration \ by \ parts]$

$\int_{}^{}f(ax)dx=\frac{1}{a}.\int_{}^{}f(u)du$

$\int_{}^{}F[f(x)]dx=\int_{}^{}[F(u)\frac{dx}{du}]du=\int_{}^{}\frac{F(u)}{f'(x)}du$ where $u = f(x)$.

$\int_{}^{}u^ndu=\frac{u^{n+1}}{n+1}$; n ≠ -1.

$\int_{}^{}\frac{du}{u}=lnu=ln|u|$;          if $u > 0$ or $\ln(-u)$ if $u < 0$.

$\int e^u \ du = e^u$

$\int_{}^{}a^udu=\int_{}^{}e^{u.lna}du=\int_{}^{}\frac{e^{u.lna}}{lna}=\frac{a^u}{lna}$;a > 0, a ≠ 1.

$\int \sin u \ du = - \cos u$

$\int \cos u \ du = \sin u \\ \int \tan u \ du = \ln \sec u = - \ln \cos u$

$\int \cot u \ du = \ln \sin u \\ \int \sec u \ du = \ln (\sec u + \tan u) = \ln \tan\frac{2u + \pi}{4}$

$\int csc u \ du = \ln (csc \ u - \cot u) = \ln \tan \frac{u}{2}$

$\int \sec^2 u \ du = \tan u \\ \int csc^2 \ u \ du = - \cot u \ du$

$\int tan^2 u \ du = \tan u - u$

$\int \cot^2 u \ du = - \cot u - u$

$\int_{}^{}sin^2udu=\frac{u}{2}-\frac{sin2u}{4}=\frac{u-sinu.cosu}{2}$

$\int_{}^{}cos^2udu=\frac{u}{2}+\frac{sin2u}{4}=\frac{u+sinu.cosu}{2}$

$\int \sec u \tan u \ du = \sec u$

$\int csc \ u \cot u \ du = - csc \ u$

$\int \sinh u \ du = \cosh u \\ \int \cosh u \ du = \sinh u$

$\int \tanh u \ du = \ln \cosh u$

$\int \coth u \ du = \ln \sinh u \\ \int sech \ u \ du = \sin^{-1}(\tanh u) \ or \ 2 \tan^{-1}.e^u$

$\int csch \ u \ du = \ln \tanh \frac{u}{2} \ or \ - \coth^{-1}.e^u$

$\int sech^2 \ u \ du = \tanh u$

$\int csch^2 \ u \ du = - \coth u$

$\int tanh^2 u \ du = u - \tanh u$

$\int \coth^2 u \ du = u - \coth u$

$\int_{}^{}sinh^2udu=\frac{sinh2u}{4}-\frac{u}{2}=\frac{sinhu.coshu-u}{2}$

$\int_{}^{}cosh^2udu=\frac{sinh2u}{4}+\frac{u}{2}=\frac{sinhu.coshu+u}{2}$

$\int sech u \tanh u \ du = - sech \ u \\ \int csch \ u \coth u \ du = - csch \ u$

$\int\frac{du}{u^2+a^2}=\frac{1}{a}\tan^{-1}\frac{u}{a}$

$\int\frac{du}{u^2-a^2}=\frac{1}{2a}\ln\left(\frac{u-a}{u+a}\right)=-\frac{1}{a}\coth^{-1}\frac{u}{a}\quad u^2>a^2$

$\int\frac{du}{a^2-u^2}=\frac{1}{2a}\ln\left(\frac{a+u}{a-u}\right)=\frac{1}{a}\tanh^{-1}\frac{u}{a}\quad\ \ \ u^2< a^2$

$\int\frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\frac{u}{a}$

$\int\frac{du}{\sqrt{u^2+a^2}}=\ln(u+\sqrt{u^2+a^2})\quad\text{or}\quad \sinh^{-1}\frac{u}{a}$

$\int\frac{du}{\sqrt{u^2-a^2}}=\ln(u+\sqrt{u^2-a^2})$

$\int\frac{du}{u\sqrt{u^2-a^2}}=\frac{1}{a}\sec^{-1}\left|\frac{u}{a}\right|$

$\int\frac{du}{u\sqrt{u^2+a^2}}=-\frac{1}{a}\ln\left(\frac{a+\sqrt{u^2+a^2}}{u}\right)$

$\int\frac{du}{u\sqrt{a^2-u^2}}=-\frac{1}{a}\ln\left(\frac{a+\sqrt{a^2-u^2}}{u}\right)$

$\int f^{(n)}g dx=f^{(n-1)}g-f^{(n-2)}g'+f^{(n-3)}g''-\cdots (-1)^n\int fg^{(n)} dx$

This is called generalized integration by parts.

Important Transformations

Often in practice an integral can be simplified by using an appropriate transformation or substitution and formula 14.6. The following list gives some transformations and their effects.

$\int_{}^{}F(ax+b)dx=\frac{1}{a}.\int_{}^{}F(u)du$          where $u = ax + b$

$\int_{}^{}F(\sqrt{ax+b})dx=\frac{2}{a}.\int_{}^{}u.F(u)du$          where $u = \sqrt{ax + b}$

$\int F(\sqrt[n]{ax+b})dx=\frac{n}{a}\int u^{n-1}F(u)du\qquad\text{where}\quad u=\sqrt[n]{ax+b}$

$\int F(\sqrt{a^2 - x^2}) \ dx = a \int F(a.\cos u). \cos u \ du \ \ \ where \ x = a \sin u$

$\int F(\sqrt{x^2 + a^2}) \ dx = a \int F(a.\sec u). \sec^2 u \ du \ \ \ where \ x = a \tan u$

$\int F(\sqrt{x^2 - a^2}) \ dx = a \int F(a.\tan u). \sec u . \tan u \ du \ \ \ where \ x = a \sec u$

$\int_{}^{}F(e^{ax})dx=\frac{1}{a}.\int_{}^{}\frac{F(u)}{u}du$          where u = eax

$\int F(\ln x) \ dx = \int F(u). e^u \ du \ \ \ where \ u = \ln x \\ \int F(\sin^{-1} \frac{x}{a}) \ dx = a \int F(u). \cos u \ du \ \ \ where \ u = \sin^{-1} \frac{x}{a}$

Similar results apply for other inverse trigonometric functions.

$\int F(\sin x, \cos x) \ dx = 2$

$\int F\left(\frac{2u}{1+u^2},\frac{1-u^2}{1+u^2}\right)\frac{du}{1+u^2}$

Special Integrals

Integrals Involving $ax + b$

$\int\frac{dx}{ax+b}=\frac{1}{a}\ln(ax+b)$

$\int\frac{xdx}{ax+b}=\frac{x}{a}-\frac{b}{a^2}\ln(ax+b)$

$\int\frac{x^2dx}{ax+b}=\frac{(ax+b)^2}{2a^3}-\frac{2b(ax+b)}{a^3}+\frac{b^2}{a^3}\ln(ax+b)$

$\int\frac{x^3dx}{ax+b}=\frac{(ax+b)^3}{3a^4}-\frac{3b(ax+b)^2}{2a^4}+\frac{3b^2(ax+b)}{a^4}-\frac{b^3}{a^4}\ln(ax+b)$

$\int\frac{dx}{x(ax+b)}=\frac{1}{b}\ln\left(\frac{x}{ax+b}\right)$

$\int\frac{dx}{x^2(ax+b)}=-\frac{1}{bx}+\frac{a}{b^2}\ln\left(\frac{ax+b}{x}\right)$

$\int\frac{dx}{x^3(ax+b)}=\frac{2ax-b}{2b^2x^2}+\frac{a^2}{b^3}\ln\left(\frac{x}{ax+b}\right)$

$\int\frac{dx}{(ax+b)^2}=\frac{-1}{a(ax+b)}$

$\int\frac{xdx}{(ax+b)^2}=\frac{b}{a^2(ax+b)}+\frac{1}{a^2}\ln(ax+b)$

$\int\frac{x^2dx}{(ax+b)^2}=\frac{ax+b}{a^3}-\frac{b^2}{a^3(ax+b)}-\frac{2b}{a^3}\ln(ax+b)$

$\int\frac{x^3dx}{(ax+b)^2}=\frac{(ax+b)^2}{2a^4}-\frac{3b(ax+b)}{a^4}+\frac{b^3}{a^4(ax+b)}+\frac{3b^2}{a^4}\ln(ax+b)$

$\int\frac{dx}{x(ax+b)^2}=\frac{1}{b(ax+b)}+\frac{1}{b^2}\ln\left(\frac{x}{ax+b}\right)$

$\int\frac{dx}{x^2(ax+b)^2}=\frac{-a}{b^2(ax+b)}-\frac{1}{b^2x}+\frac{2a}{b^3}\ln\left(\frac{ax+b}{x}\right)$

$\int\frac{dx}{x^3(ax+b)^2}=-\frac{(ax+b)^2}{2b^4x^2}+\frac{3a(ax+b)}{b^4x}-\frac{a^3x}{b^4(ax+b)}-\frac{3a^2}{b^4}\ln\left(\frac{ax+b}{x}\right)$

$\int\frac{dx}{(ax+b)^3}=\frac{-1}{2(ax+b)^2}$

$\int\frac{xdx}{(ax+b)^3}=\frac{-1}{a^2(ax+b)}+\frac{b}{2a^2(ax+b)^2}$

$\int\frac{x^2dx}{(ax+b)^3}=\frac{2b}{a^3(ax+b)}-\frac{b^2}{2a^3(ax+b)^2}+\frac{1}{a^3}\ln(ax+b)$

$\int\frac{x^3dx}{(ax+b)^3}=\frac{x}{a^3}-\frac{3b^2}{a^4(ax+b)}+\frac{b^3}{2a^4(ax+b)^2}-\frac{3b}{a^4}\ln(ax+b)$

$\int\frac{dx}{x(ax+b)^3}=\frac{a^2x^2}{2b^3(ax+b)^2}-\frac{2ax}{b^3(ax+b)}-\frac{1}{b^3}\ln\left(\frac{ax+b}{x}\right)$

$\int\frac{dx}{x^2(ax+b)^3}=\frac{-a}{2b^2(ax+b)^2}-\frac{2a}{b^3(ax+b)}-\frac{1}{b^3x}+\frac{3a}{b^4}\ln\left(\frac{ax+b}{x}\right)$

$\int\frac{dx}{x^3(ax+b)^3}=-\frac{a^4x^2}{2b^5(ax+b)^2}-\frac{4a^3x}{b^5(ax+b)}-\frac{(ax+b)^2}{2b^5x^2}-\frac{6a^2}{b^5}\ln\left(\frac{ax+b}{x}\right)$

$\int(ax+b)^n dx=\frac{(ax+b)^{n+1}}{(n+1)a}.$

$\int x(ax+b)^n dx=\frac{(ax+b)^{n+2}}{(n+2)a^2}-\frac{b(ax+b)^{n+1}}{(n+1)a^2},\quad\ n\neq-1,-2$

$\int x^2(ax+b)^n dx=\frac{(ax+b)^{n+3}}{(n+3)a^3}-\frac{2b(ax+b)^{n+2}}{(n+2)a^3}+\frac{b^2(ax+b)^{n+1}}{(n+1)a^3}$

$\int x^m(ax+b)^n dx=\left\{\begin{array}{lr}\frac{x^{m+1}(ax+b)^n}{m+n+1}+\frac{nb}{m+n+1}\int x^m(ax+b)^{n-1} dx \\ \frac{x^m(ax+b)^{n+1}}{(m+n+1)a}-\frac{mb}{(m+n+1)a}\int x^{m-1}(ax+b)^n dx \\ \frac{-x^{m+1}(ax+b)^{n+1}}{(n+1)b}+\frac{m+n+2}{(n+1)b}\int x^m(ax+b)^{n+1} dx\end{array}\right.$

Integrals Involving $\sqrt{ax + b}$

$\int\frac{dx}{\sqrt{ax+b}}=\frac{2\sqrt{ax+b}}{a}$

$\int\frac{xdx}{\sqrt{ax+b}}=\frac{2(ax-2b)}{3a^2}\sqrt{ax+b}$

$\int\frac{x^2dx}{\sqrt{ax+b}}=\frac{2(3a^2x^2-4abx+8b^2)}{15a^3}\sqrt{ax+b}$

$\int\frac{dx}{x\sqrt{ax+b}}=\left\{\begin{array}{lr}\frac{1}{\sqrt{b}}\ln\left(\frac{\sqrt{ax+b}-\sqrt{b}}{\sqrt{ax+b}+\sqrt{b}}\right) \\ \frac{2}{\sqrt{-b}}\tan^{-1}\sqrt{\frac{ax+b}{-b}}\end{array}\right.$

$\int\frac{dx}{x^2\sqrt{ax+b}}=-\frac{\sqrt{ax+b}}{bx}-\frac{a}{2b}\int\frac{dx}{x\sqrt{ax+b}}$

$\int\sqrt{ax+b}\ dx=\frac{2\sqrt{(ax+b)^3}}{3a}$

$\int x\sqrt{ax+b}\ dx=\frac{2(3ax-2b)}{15a^2}\sqrt{(ax+b)^3}$

$\int x^2\sqrt{ax+b}\ dx=\frac{2(15a^2x^2-12abx+8b^2)}{105a^3}\sqrt{(ax+b)^3}$

$\int\frac{\sqrt{ax+b}}{x}dx=2\sqrt{ax+b}+b\int\frac{dx}{x\sqrt{ax+b}}$

$\int\frac{\sqrt{ax+b}}{x^2}dx=-\frac{\sqrt{ax+b}}{x}+\frac{a}{2}\int\frac{dx}{x\sqrt{ax+b}}$

$\int\frac{x^m}{\sqrt{ax+b}}dx=\frac{2x^m\sqrt{ax+b}}{(2m+1)a}-\frac{2mb}{(2m+1)a}\int\frac{x^{m-1}}{\sqrt{ax+b}}dx$

$\int\frac{dx}{x^m\sqrt{ax+b}}=-\frac{\sqrt{ax+b}}{(m-1)bx^{m-1}}-\frac{(2m-3)a}{(2m-2)b}\int\frac{dx}{x^{m-1}\sqrt{ax+b}}$

$\int x^m\sqrt{ax+b}\ dx=\frac{2x^m}{(2m+3)a}(ax+b)^\frac{3}{2}-\frac{2mb}{(2m+3)a}\int x^{m-1}\sqrt{ax+b}\ dx$

$\int\frac{\sqrt{ax+b}}{x^m}dx=-\frac{\sqrt{ax+b}}{(m-1)x^{m-1}}+\frac{a}{2(m-1)}\int\frac{dx}{x^{m-1}\sqrt{ax+b}}$

$\int\frac{\sqrt{ax+b}}{x^m}dx=\frac{-(ax+b)^\frac{3}{2}}{(m-1)bx^{m-1}}-\frac{(2m-5)a}{(2m-2)b}\int\frac{\sqrt{ax+b}}{x^{m-1}}dx$

$\int(ax+b)^\frac{m}{2}dx=\frac{2(ax+b)^\frac{(m+2)}{2}}{a(m+2)}$

$\int x(ax+b)^\frac{m}{2}dx=\frac{2(ax+b)^\frac{(m+4)}{2}}{a^2(m+4)}-\frac{2b(ax+b)^\frac{(m+2)}{2}}{a^2(m+2)}$

$\int x^2(ax+b)^\frac{m}{2}dx=\frac{2(ax+b)^\frac{(m+6)}{2}}{a^3(m+6)}-\frac{4b(ax+b)^\frac{(m+4)}{2}}{a^3(m+4)}+\frac{2b^2(ax+b)^\frac{(m+2)}{2}}{a^3(m+2)}$

$\int\frac{(ax+b)^\frac{m}{2}}{x}dx=\frac{2(ax+b)^\frac{m}{2}}{m}+b\int\frac{(ax+b)^\frac{(m-2)}{2}}{x}dx$

$\int\frac{(ax+b)^\frac{m}{2}}{x^2}dx=-\frac{(ax+b)^\frac{(m+2)}{2}}{bx}+\frac{ma}{2b}\int\frac{(ax+b)^\frac{m}{2}}{x}dx$

$\int\frac{dx}{x(ax+b)^\frac{m}{2}}=\frac{2}{(m-2)b(ax+b)^\frac{(m-2)}{2}}+\frac{1}{b}\int\frac{dx}{x(ax+b)^\frac{(m-2)}{2}}$

Integrals Involving $ax + b$ and $px + q$

$\int\frac{dx}{(ax+b)(px+q)}=\frac{1}{bp-aq}\ln\left(\frac{px+q}{ax+b}\right)$

$\int\frac{xdx}{(ax+b)(px+q)}=\frac{1}{bp-aq}\left\{\frac{b}{a}\ln(ax+b)-\frac{q}{p}\ln(px+q)\right\}$

$\int\frac{dx}{(ax+b)^2(px+q)}=\frac{1}{bp-aq}\left\{\frac{1}{ax+b}+\frac{p}{bp-aq}\ln\left(\frac{px+q}{ax+b}\right)\right\}$

$\int\frac{xdx}{(ax+b)^2(px+q)}=\frac{1}{bp-aq}\left\{\frac{q}{bp-aq}\ln\left(\frac{ax+b}{px+q}\right)-\frac{b}{a(ax+b)}\right\}$

$\int\frac{x^2dx}{(ax+b)^2(px+q)}=\frac{b^2}{(bp-aq)a^2(ax+b)}+\frac{1}{(bp-aq)^2}\left\{\frac{q^2}{p}\ln(px+q)+\frac{b(bp-2aq)}{a^2}\ln(ax+b)\right\}$

$\int\frac{dx}{(ax+b)^m(px+q)^n}=\frac{-1}{(n-1)(bp-aq)}\left\{\frac{1}{(ax+b)^{m-1}(px+q)^{n-1}}+a(m+n-2)\int\frac{dx}{(ax+b)^m(px+q)^{n-1}}\right\}$

$\int\frac{ax+b}{px+q}dx=\frac{ax}{p}+\frac{bp-aq}{p^2}\ln(px+q)$

$\int\frac{(ax+b)^m}{(px+q)^n} dx=\left\{\begin{array}{lr}\frac{-1}{(n-1)(bp-aq)}\left\{\frac{(ax+b)^{m+1}}{(px+q)^{n-1}}+(n-m-2)a\int\frac{(ax+b)^m}{(px+q)^{n-1}} dx\right\} \\ \frac{-1}{(n-m-1)p}\left\{\frac{(ax+b)^m}{(px+q)^{n-1}}+m(bp-aq)\int\frac{(ax+b)^{m-1}}{(px+q)^n} dx\right\} \\ \frac{-1}{(n-1)p}\left\{\frac{(ax+b)^m}{(px+q)^{n-1}}+ma\int\frac{(ax+b)^{m-1}}{(px+q)^{n-1}} dx\right\}\end{array}\right.$

Integrals Involving $\sqrt{ax + b}$ and $px + q$

$\int\frac{px+q}{\sqrt{ax+b}} dx=\frac{2(apx+3aq-2bp)}{3a^2}\sqrt{ax+b}$

$\int\frac{dx}{(px+q)\sqrt{ax+b}}=\left\{\begin{array}{lr}\frac{1}{\sqrt{bp-aq}\sqrt{p}}\ln\left(\frac{\sqrt{p(ax+b)}-\sqrt{bp-aq}}{\sqrt{p(ax+b)}+\sqrt{bp-aq}}\right) \\ \frac{2}{\sqrt{aq-bp}\sqrt{p}}\tan^{-1}\sqrt{\frac{p(ax+b)}{aq-bp}}\end{array}\right.$

$\int\frac{\sqrt{ax+b}}{px+q} dx=\left\{\begin{array}{lr}\frac{2\sqrt{ax+b}}{p}+\frac{\sqrt{bp-aq}}{p\sqrt{p}}\ln\left(\frac{\sqrt{p(ax+b)}-\sqrt{bp-aq}}{\sqrt{p(ax+b)}+\sqrt{bp-aq}}\right) \\ \frac{2\sqrt{ax+b}}{p}-\frac{2\sqrt{aq-bp}}{p\sqrt{p}}\tan^{-1}\sqrt{\frac{p(ax+b)}{aq-bp}}\end{array}\right.$

$\int(px+q)^n\sqrt{ax+b}\ dx=\frac{2(px+q)^{n+1}\sqrt{ax+b}}{(2n+3)p}+\frac{bp-aq}{(2n+3)p}\int\frac{(px+q)^n}{\sqrt{ax+b}} dx$

$\int\frac{dx}{(px+q)^n\sqrt{ax+b}}=\frac{\sqrt{ax+b}}{(n-1)(aq-bp)(px+q)^{n-1}}+\frac{(2n-3)a}{2(n-1)(aq-bp)}\int\frac{dx}{(px+q)^{n-1}\sqrt{ax+b}}$

$\int\frac{(px+q)^n}{\sqrt{ax+b}}\ dx=\frac{2(px+q)^n\sqrt{ax+b}}{(2n+1)a}+\frac{2n(aq-bp)}{(2n+1)a}\int\frac{(px+q)^{n-1}\ dx}{\sqrt{ax+b}}$

$\int\frac{\sqrt{ax+b}}{(px+q)^n}\ dx=\frac{-\sqrt{ax+b}}{(n-1)p(px+q)^{n-1}}+\frac{a}{2(n-1)p}\int\frac{dx}{(px+q)^{n-1}\sqrt{ax+b}}$

Integrals Involving $\sqrt{ax+b}$ and $\sqrt{px+q}$

$\int\frac{dx}{\sqrt{(ax+b)(px+q)}}=\left\{\begin{array}{lr}\frac{2}{\sqrt{ap}}\ln\left(\sqrt{a(px+q)}+\sqrt{p(ax+b)}\right) \\ -\frac{2}{\sqrt{-ap}}\tan^{-1}\sqrt{\frac{-p(ax+b)}{a(px+q)}}\end{array}\right.$

$\int\frac{x\ dx}{\sqrt{(ax+b)(px+q)}}=\frac{\sqrt{(ax+b)(px+q)}}{ap}-\frac{bp+aq}{2ap}\int\frac{dx}{\sqrt{(ax+b)(px+q)}}$

$\int\sqrt{(ax+b)(px+q)}\ dx=\frac{2apx+bp+aq}{4ap}\sqrt{(ax+b)(px+q)}-\frac{(bp-aq)^2}{8ap}\int\frac{dx}{\sqrt{(ax+b)(px+q)}}$

$\int\sqrt{\frac{px+q}{ax+b}}\ dx=\frac{\sqrt{(ax+b)(px+q)}}{a}+\frac{aq-bp}{2a}\int\frac{dx}{\sqrt{(ax+b)(px+q)}}$

$\int\frac{dx}{(px+q)\sqrt{(ax+b)(px+q)}}=\frac{2\sqrt{ax+b}}{(aq-bp)\sqrt{px+q}}$

Integrals Involving $x^2+a^2$

$\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\frac{x}{a}$

$\int\frac{x\ dx}{x^2+a^2}=\frac{1}{2}\ln(x^2+a^2)$

$\int\frac{x^2\ dx}{x^2+a^2}=x-a\tan^{-1}\frac{x}{a}$

$\int\frac{x^3\ dx}{x^2+a^2}=\frac{x^2}{2}-\frac{a^2}{2}\ln(x^2+a^2)$

$\int\frac{dx}{x(x^2+a^2)}=\frac{1}{2a^2}\ln\left(\frac{x^2}{x^2+a^2}\right)$

$\int\frac{dx}{x^2(x^2+a^2)}=-\frac{1}{a^2x}-\frac{1}{a^3}\tan^{-1}\frac{x}{a}$

$\int\frac{dx}{x^3(x^2+a^2)}=-\frac{1}{2a^2x^2}-\frac{1}{2a^4}\ln\left(\frac{x^2}{x^2+a^2}\right)$

$\int\frac{dx}{(x^2+a^2)^2}=\frac{x}{2a^2(x^2+a^2)}+\frac{1}{2a^3}\tan^{-1}\frac{x}{a}$

$\int\frac{x\ dx}{(x^2+a^2)^2}=\frac{1}{2(x^2+a^2)}$

$\int\frac{x^2\ dx}{(x^2+a^2)^2}=\frac{-x}{2(x^2+a^2)}+\frac{1}{2a}\tan^{-1}\frac{x}{a}$

$\int\frac{x^3\ dx}{(x^2+a^2)^2}=\frac{a^2}{2(x^2+a^2)}+\frac{1}{2}\ln(x^2+a^2)$

$\int\frac{dx}{x(x^2+a^2)^2}=\frac{1}{2a^2(x^2+a^2)}+\frac{1}{2a^4}\ln\left(\frac{x^2}{x^2+a^2}\right)$

$\int\frac{dx}{x^2(x^2+a^2)^2}=-\frac{1}{a^4x}-\frac{x}{2a^4(x^2+a^2)}-\frac{3}{2a^5}\tan^{-1}\frac{x}{a}$

$\int\frac{dx}{x^3(x^2+a^2)^2}=-\frac{1}{2a^4x^2}-\frac{1}{2a^4(x^2+a^2)}-\frac{1}{a^6}\ln\left(\frac{x^2}{x^2+a^2}\right)$

$\int\frac{dx}{(x^2+a^2)^n}=\frac{x}{2(n-1)a^2(x^2+a^2)^{n-1}}+\frac{2n-3}{(2n-2)a^2}\int\frac{dx}{(x^2+a^2)^{n-1}}$

$\int\frac{x\ dx}{(x^2+a^2)^n}=\frac{-1}{2(n-1)(x^2+a^2)^{n-1}}$

$\int\frac{dx}{x(x^2+a^2)^n}=\frac{1}{2(n-1)a^2(x^2+a^2)^{n-1}}+\frac{1}{a^2}\int\frac{dx}{x(x^2+a^2)^{n-1}}$

$\int\frac{x^m\ dx}{(x^2+a^2)^n}=\int\frac{x^{m-2}dx}{(x^2+a^2)^{n-1}}-a^2\int\frac{x^{m-2}dx}{(x^2+a^2)^n}$

$\int\frac{dx}{x^m(x^2+a^2)^n}=\frac{1}{a^2}\int\frac{dx}{x^m(x^2+a^2)^{n-1}}-\frac{1}{a^2}\int\frac{dx}{x^{m-2}(x^2+a^2)^n}$

Integrals Involving $x^2-a^2,\quad x^2>a^2$

$\int\frac{dx}{x^2-a^2}=\frac{1}{2a}\ln\left(\frac{x-a}{x+a}\right)\quad\text{or}\quad -\frac{1}{a}\coth^{-1}\frac{x}{a}$

$\int\frac{x\ dx}{x^2-a^2}=\frac{1}{2}\ln(x^2-a^2)$

$\int\frac{x^2dx}{x^2-a^2}=x+\frac{a}{2}\ln\left(\frac{x-a}{x+a}\right)$

$\int\frac{x^3 dx}{x^2-a^2}=\frac{x^2}{2}+\frac{a^2}{2}\ln(x^2-a^2)$

$\int\frac{dx}{x(x^2-a^2)}=\frac{1}{2a^2}\ln\left(\frac{x^2-a^2}{x^2}\right)$

$\int\frac{dx}{x^2(x^2-a^2)}=\frac{1}{a^2x}+\frac{1}{2a^3}\ln\left(\frac{x-a}{x+a}\right)$

$\int\frac{dx}{x^3(x^2-a^2)}=\frac{1}{2a^2x^2}-\frac{1}{2a^4}\ln\left(\frac{x^2}{x^2-a^2}\right)$

$\int\frac{dx}{(x^2-a^2)^2}=\frac{-x}{2a^2(x^2-a^2)}-\frac{1}{4a^3}\ln\left(\frac{x-a}{x+a}\right)$

$\int\frac{x\ dx}{(x^2-a^2)^2}=\frac{-1}{2(x^2-a^2)}$

$\int\frac{x^2dx}{(x^2-a^2)^2}=\frac{-x}{2(x^2-a^2)}+\frac{1}{4a}\ln\left(\frac{x-a}{x+a}\right)$

$\int\frac{x^3dx}{(x^2-a^2)^2}=\frac{-a^2}{2(x^2-a^2)}+\frac{1}{2}\ln(x^2-a^2)$

$\int\frac{dx}{x(x^2-a^2)^2}=\frac{-1}{2a^2(x^2-a^2)}+\frac{1}{2a^4}\ln\left(\frac{x^2}{x^2-a^2}\right)$

$\int\frac{dx}{x^2(x^2-a^2)^2}=-\frac{1}{a^4x}-\frac{x}{2a^4(x^2-a^2)}-\frac{3}{4a^5}\ln\left(\frac{x-a}{x+a}\right)$

$\int\frac{dx}{x^3(x^2-a^2)^2}=-\frac{1}{2a^4x^2}-\frac{1}{2a^4(x^2-a^2)}+\frac{1}{a^6}\ln\left(\frac{x^2}{x^2-a^2}\right)$

$\int\frac{dx}{(x^2-a^2)^n}=\frac{-x}{2(n-1)a^2(x^2-a^2)^{n-1}}-\frac{2n-3}{(2n-2)a^2}\int\frac{dx}{(x^2-a^2)^{n-1}}$

$\int\frac{x\ dx}{(x^2-a^2)^n}=\frac{-1}{2(n-1)(x^2-a^2)^{n-1}}$

$\int\frac{dx}{x(x^2-a^2)^n}=\frac{-1}{2(n-1)a^2(x^2-a^2)^{n-1}}-\frac{1}{a^2}\int\frac{dx}{x(x^2-a^2)^{n-1}}$

$\int\frac{x^m dx}{(x^2-a^2)^n}=\int\frac{x^{m-2}dx}{(x^2-a^2)^{n-1}}+a^2\int\frac{x^{m-2}dx}{(x^2-a^2)^n}$

$\int\frac{dx}{x^m(x^2-a^2)^n}=\frac{1}{a^2}\int\frac{dx}{x^{m-2}(x^2-a^2)^n}-\frac{1}{a^2}\int\frac{dx}{x^m(x^2-a^2)^{n-1}}$

Integrals Involving $a^2-x^2,\quad x^2< a^2$

$\int\frac{dx}{a^2-x^2}=\frac{1}{2a}\ln\left(\frac{a+x}{a-x}\right)\quad\text{or}\quad \frac{1}{a}\tanh^{-1}\frac{x}{a}$

$\int\frac{x\ dx}{a^2-x^2}=-\frac{1}{2}\ln(a^2-x^2)$

$\int\frac{x^2dx}{a^2-x^2}=-x+\frac{a}{2}\ln\left(\frac{a+x}{a-x}\right)$

$\int\frac{x^3 dx}{a^2-x^2}=-\frac{x^2}{2}-\frac{a^2}{2}\ln(a^2-x^2)$

$\int\frac{dx}{x(a^2-x^2)}=\frac{1}{2a^2}\ln\left(\frac{x^2}{a^2-x^2}\right)$

$\int\frac{dx}{x^2(a^2-x^2)}=-\frac{1}{a^2x}+\frac{1}{2a^3}\ln\left(\frac{a+x}{a-x}\right)$

$\int\frac{dx}{x^3(a^2-x^2)}=-\frac{1}{2a^2x^2}+\frac{1}{2a^4}\ln\left(\frac{x^2}{a^2-x^2}\right)$

$\int\frac{dx}{(a^2-x^2)^2}=\frac{x}{2a^2(a^2-x^2)}+\frac{1}{4a^3}\ln\left(\frac{a+x}{a-x}\right)$

$\int\frac{x\ dx}{(a^2-x^2)^2}=\frac{1}{2(a^2-x^2)}$

$\int\frac{x^2dx}{(a^2-x^2)^2}=\frac{x}{2(a^2-x^2)}-\frac{1}{4a}\ln\left(\frac{a+x}{a-x}\right)$

$\int\frac{x^3dx}{(a^2-x^2)^2}=\frac{a^2}{2(a^2-x^2)}+\frac{1}{2}\ln(a^2-x^2)$

$\int\frac{dx}{x(a^2-x^2)^2}=\frac{1}{2a^2(a^2-x^2)}+\frac{1}{2a^4}\ln\left(\frac{x^2}{a^2-x^2}\right)$

$\int\frac{dx}{x^2(a^2-x^2)^2}=-\frac{1}{a^4x}+\frac{x}{2a^4(a^2-x^2)}+\frac{3}{4a^5}\ln\left(\frac{a+x}{a-x}\right)$

$\int\frac{dx}{x^3(a^2-x^2)^2}=\frac{-1}{2a^4x^2}+\frac{1}{2a^4(a^2-x^2)}+\frac{1}{a^6}\ln\left(\frac{x^2}{a^2-x^2}\right)$

$\int\frac{dx}{(a^2-x^2)^n}=\frac{x}{2(n-1)a^2(a^2-x^2)^{n-1}}+\frac{2n-3}{(2n-2)a^2}\int\frac{dx}{(a^2-x^2)^{n-1}}$

$\int\frac{x\ dx}{(a^2-x^2)^n}=\frac{1}{2(n-1)(a^2-x^2)^{n-1}}$

$\int\frac{dx}{x(a^2-x^2)^n}=\frac{1}{2(n-1)a^2(a^2-x^2)^{n-1}}+\frac{1}{a^2}\int\frac{dx}{x(a^2-x^2)^{n-1}}$

$\int\frac{x^m dx}{(a^2-x^2)^n}=a^2\int\frac{x^{m-2}dx}{(a^2-x^2)^n}-\int\frac{x^{m-2}dx}{(a^2-x^2)^{n-1}}$

$\int\frac{dx}{x^m(a^2-x^2)^n}=\frac{1}{a^2}\int\frac{dx}{x^m(a^2-x^2)^{n-1}}+\frac{1}{a^2}\int\frac{dx}{x^{m-2}(a^2-x^2)^n}$

Integrals Involving $\sqrt{x^2+a^2}$

$\int\frac{dx}{\sqrt{x^2+a^2}}=\ln(x+\sqrt{x^2+a^2})\quad\text{or}\quad\sinh^{-1}\frac{x}{a}$

$\int\frac{x\ dx}{\sqrt{x^2+a^2}}=\sqrt{x^2+a^2}$

$\int\frac{x^2 dx}{\sqrt{x^2+a^2}}=\frac{x\sqrt{x^2+a^2}}{2}-\frac{a^2}{2}\ln(x+\sqrt{x^2+a^2})$

$\int\frac{x^3 dx}{\sqrt{x^2+a^2}}=\frac{(x^2+a^2)^\frac{3}{2}}{3}-a^2\sqrt{x^2+a^2}$

$\int\frac{dx}{x\sqrt{x^2+a^2}}=-\frac{1}{a}\ln\left(\frac{a+\sqrt{x^2+a^2}}{x}\right)$

$\int\frac{dx}{x^2\sqrt{x^2+a^2}}=-\frac{\sqrt{x^2+a^2}}{a^2x}$

$\int\frac{dx}{x^3\sqrt{x^2+a^2}}=-\frac{\sqrt{x^2+a^2}}{2a^2x^2}+\frac{1}{2a^3}\ln\left(\frac{a+\sqrt{x^2+a^2}}{x}\right)$

$\int\sqrt{x^2+a^2}\ dx=\frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2}{2}\ln(x+\sqrt{x^2+a^2})$

$\int x\sqrt{x^2+a^2}\ dx=\frac{(x^2+a^2)^\frac{3}{2}}{3}$

$\int x^2\sqrt{x^2+a^2}\ dx=\frac{x(x^2+a^2)^\frac{3}{2}}{4}-\frac{a^2x\sqrt{x^2+a^2}}{8}-\frac{a^4}{8}\ln(x+\sqrt{x^2+a^2})$

$\int x^3\sqrt{x^2+a^2}\ dx=\frac{(x^2+a^2)^\frac{5}{2}}{5}-\frac{a^2(x^2+a^2)^\frac{3}{2}}{3}$

$\int\frac{\sqrt{x^2+a^2}}{x} dx=\sqrt{x^2+a^2}-a\ln\left(\frac{a+\sqrt{x^2+a^2}}{x}\right)$

$\int\frac{\sqrt{x^2+a^2}}{x^2} dx=-\frac{\sqrt{x^2+a^2}}{x}+\ln(x+\sqrt{x^2+a^2})$

$\int\frac{\sqrt{x^2+a^2}}{x^3} dx=-\frac{\sqrt{x^2+a^2}}{2x^2}-\frac{1}{2a}\ln\left(\frac{a+\sqrt{x^2+a^2}}{x}\right)$

$\int\frac{dx}{(x^2+a^2)^\frac{3}{2}}=\frac{x}{a^2\sqrt{x^2+a^2}}$

$\int\frac{x\ dx}{(x^2+a^2)^\frac{3}{2}}=\frac{-1}{\sqrt{x^2+a^2}}$

$\int\frac{x^2dx}{(x^2+a^2)^\frac{3}{2}}=\frac{-x}{\sqrt{x^2+a^2}}+\ln(x+\sqrt{x^2+a^2})$

$\int\frac{x^3dx}{(x^2+a^2)^\frac{3}{2}}=\sqrt{x^2+a^2}+\frac{a^2}{\sqrt{x^2+a^2}}$

$\int\frac{dx}{x(x^2+a^2)^\frac{3}{2}}=\frac{1}{a^2\sqrt{x^2+a^2}}-\frac{1}{a^3}\ln\left(\frac{a+\sqrt{x^2+a^2}}{x}\right)$

$\int\frac{dx}{x^2(x^2+a^2)^\frac{3}{2}}=-\frac{\sqrt{x^2+a^2}}{a^4x}-\frac{x}{a^4\sqrt{x^2+a^2}}$

$\int\frac{dx}{x^3(x^2+a^2)^\frac{3}{2}}=\frac{-1}{2a^2x^2\sqrt{x^2+a^2}}-\frac{3}{2a^4\sqrt{x^2+a^2}}+\frac{3}{2a^5}\ln\left(\frac{a+\sqrt{x^2+a^2}}{x}\right)$

$\int(x^2+a^2)^\frac{3}{2}\ dx=\frac{x(x^2+a^2)^\frac{3}{2}}{4}+\frac{3a^2x\sqrt{x^2+a^2}}{8}+\frac{3}{8}a^4\ln(x+\sqrt{x^2+a^2})\int x(x^2+a^2)^\frac{3}{2}\ dx=\frac{(x^2+a^2)^\frac{5}{2}}{5}$

$\int x^2(x^2+a^2)^\frac{3}{2}\ dx=\frac{x(x^2+a^2)^\frac{5}{2}}{6}-\frac{a^2x(x^2+a^2)^\frac{3}{2}}{24}-\frac{a^4x\sqrt{x^2+a^2}}{16}-\frac{a^5}{16}\ln(x+\sqrt{x^2+a^2})$

$\int x^3(x^2+a^2)^\frac{3}{2}\ dx=\frac{(x^2+a^2)^\frac{7}{2}}{7}-\frac{a^2(x^2+a^2)^\frac{5}{2}}{5}$

$\int\frac{(x^2+a^2)^\frac{3}{2}}{x}\ dx=\frac{(x^2+a^2)^\frac{3}{2}}{3}+a^2\sqrt{x^2+a^2}-a^3\ln\left(\frac{a+\sqrt{x^2+a^2}}{x}\right)$

$\int\frac{(x^2+a^2)^\frac{3}{2}}{x^2}\ dx=-\frac{(x^2+a^2)^\frac{3}{2}}{x}+\frac{3x\sqrt{x^2+a^2}}{2}+\frac{3}{2}a^2\ln(x+\sqrt{x^2+a^2})$

$\int\frac{(x^2+a^2)^\frac{3}{2}}{x^3}\ dx=-\frac{(x^2+a^2)^\frac{3}{2}}{2x^2}+\frac{3}{2}\sqrt{x^2+a^2}-\frac{3}{2}a\ln\left(\frac{a+\sqrt{x^2+a^2}}{x}\right)$


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