Engineering Handbook/Calculus/Integration/irrational functions

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

$\int r^{3}dx=\frac{1}{4}x{r}^3+\frac{3}{8}{a}^{2}xr+\frac{3}{8}{a}^4\ln (x+r)$

$\int r^{5}dx=\frac{1}{6}x{r}^5+\frac{5}{24}{a}^{2}x{r}^3+\frac{5}{16}{a}^{4}xr+\frac{5}{16}{a}^6\ln (x+r)$

$\int xrdx=\frac{r^3}{3}$

$\int x{r}^{3}dx=\frac{r^5}{5}$

$\int x{r}^{2n+1}dx=\frac{r^{2n+3}}{2n+3}$

$\int x^{2}rdx=\frac{x{r}^3}{4}-\frac{a^{2}xr}{8}-\frac{a^4}{8}\ln (x+r)$

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

$\int x^{3}rdx=\frac{r^5}{5}-\frac{a^2{r}^3}{3}$

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

$\int x^3{r}^{2n+1}dx=\frac{r^{2n+5}}{2n+5}-\frac{a^2{r}^{2n+3}}{2n+3}$

$\int x^{4}rdx=\frac{x^3{r}^3}{6}-\frac{a^{2}x{r}^3}{8}+\frac{a^{4}xr}{16}+\frac{a^6}{16}\ln (x+r)$

$\int x^4{r}^{3}dx=\frac{x^3{r}^5}{8}-\frac{a^{2}x{r}^5}{16}+\frac{a^{4}x{r}^3}{64}+\frac{3a^{6}xr}{128}+\frac{3a^8}{128}\ln (x+r)$

$\int x^{5}rdx=\frac{r^7}{7}-\frac{2a^2{r}^5}{5}+\frac{a^4{r}^3}{3}$

$\int x^5{r}^{3}dx=\frac{r^9}{9}-\frac{2a^2{r}^7}{7}+\frac{a^4{r}^5}{5}$

$\int x^5{r}^{2n+1}dx=\frac{r^{2n+7}}{2n+7}-\frac{2a^2{r}^{2n+5}}{2n+5}+\frac{a^4{r}^{2n+3}}{2n+3}$

$\int \frac{rdx}{x}=r-a\ln \left|\frac{a+r}{x}\right|=r-a\mathrm{arsinh}\frac{a}{x}$

$\int \frac{r^{3}dx}{x}=\frac{r^3}{3}+a^{2}r-a^3\ln \left|\frac{a+r}{x}\right|$

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

$\int \frac{r^{7}dx}{x}=\frac{r^7}{7}+\frac{a^2{r}^5}{5}+\frac{a^4{r}^3}{3}+a^{6}r-a^7\ln \left|\frac{a+r}{x}\right|$

$\int \frac{dx}{r}=\mathrm{arsinh}\frac{x}{a}=\ln \left(\frac{x+r}{a}\right)$

$\int \frac{dx}{r^3}=\frac{x}{a^{2}r}$

$\int \frac{xdx}{r}=r$

$\int \frac{xdx}{r^3}=-\frac{1}{r}$

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

$\int \frac{dx}{xr}=-\frac{1}{a}\mathrm{arsinh}\frac{a}{x}=-\frac{1}{a}\ln \left|\frac{a+r}{x}\right|$

Integrals involving $s=\sqrt{x^2-a^2}$

Assume $(x^2>a^2)$, for $(x^2<a^2)$, see next section:

$\int sdx=\frac{1}{2}(xs-a^2\ln (x+s))$

$\int xsdx=\frac{1}{3}{s}^3$

$\int \frac{sdx}{x}=s-a\arccos \left|\frac{a}{x}\right|$

$\int \frac{dx}{s}=\ln \left|\frac{x+s}{a}\right|$

Here $\ln \left|\frac{x+s}{a}\right|=\mathrm{s}\mathrm{g}\mathrm{n}(x)\mathrm{arcosh}\left|\frac{x}{a}\right|=\frac{1}{2}\ln \left(\frac{x+s}{x-s}\right)$, where the positive value of $\mathrm{arcosh}\left|\frac{x}{a}\right|$ is to be taken.

$\int \frac{xdx}{s}=s$

$\int \frac{xdx}{s^3}=-\frac{1}{s}$

$\int \frac{xdx}{s^5}=-\frac{1}{3s^3}$

$\int \frac{xdx}{s^7}=-\frac{1}{5s^5}$

$\int \frac{xdx}{s^{2n+1}}=-\frac{1}{(2n-1)s^{2n-1}}$

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

$\int \frac{x^{2}dx}{s}=\frac{xs}{2}+\frac{a^2}{2}\ln \left|\frac{x+s}{a}\right|$

$\int \frac{x^{2}dx}{s^3}=-\frac{x}{s}+\ln \left|\frac{x+s}{a}\right|$

$\int \frac{x^{4}dx}{s}=\frac{x^{3}s}{4}+\frac{3}{8}{a}^{2}xs+\frac{3}{8}{a}^4\ln \left|\frac{x+s}{a}\right|$

$\int \frac{x^{4}dx}{s^3}=\frac{xs}{2}-\frac{a^{2}x}{s}+\frac{3}{2}{a}^2\ln \left|\frac{x+s}{a}\right|$

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

$\int \frac{x^{2m}dx}{s^{2n+1}}=(-1{)}^{n-m}\frac{1}{a^{2(n-m)}}\sum _{i=0}^{n-m-1}\frac{1}{2(m+i)+1}(\genfrac{}{}{0ex}{}{n-m-1}{i})\frac{x^{2(m+i)+1}}{s^{2(m+i)+1}}\text{(}n>m\ge 0\text{)}$

$\int \frac{dx}{s^3}=-\frac{1}{a^2}\frac{x}{s}$

$\int \frac{dx}{s^5}=\frac{1}{a^4}[\frac{x}{s}-\frac{1}{3}\frac{x^3}{s^3}]$

$\int \frac{dx}{s^7}=-\frac{1}{a^6}[\frac{x}{s}-\frac{2}{3}\frac{x^3}{s^3}+\frac{1}{5}\frac{x^5}{s^5}]$

$\int \frac{dx}{s^9}=\frac{1}{a^8}[\frac{x}{s}-\frac{3}{3}\frac{x^3}{s^3}+\frac{3}{5}\frac{x^5}{s^5}-\frac{1}{7}\frac{x^7}{s^7}]$

$\int \frac{x^{2}dx}{s^5}=-\frac{1}{a^2}\frac{x^3}{3s^3}$

$\int \frac{x^{2}dx}{s^7}=\frac{1}{a^4}[\frac{1}{3}\frac{x^3}{s^3}-\frac{1}{5}\frac{x^5}{s^5}]$

$\int \frac{x^{2}dx}{s^9}=-\frac{1}{a^6}[\frac{1}{3}\frac{x^3}{s^3}-\frac{2}{5}\frac{x^5}{s^5}+\frac{1}{7}\frac{x^7}{s^7}]$

Integrals involving $u=\sqrt{a^2-x^2}$

$\int udx=\frac{1}{2}(xu+a^2\arcsin \frac{x}{a})\text{(}|x|\le |a|\text{)}$

$\int xudx=-\frac{1}{3}{u}^3\text{(}|x|\le |a|\text{)}$

$\int x^{2}udx=-\frac{x}{4}{u}^3+\frac{a^2}{8}(xu+a^2\arcsin \frac{x}{a})\text{(}|x|\le |a|\text{)}$

$\int \frac{udx}{x}=u-a\ln \left|\frac{a+u}{x}\right|\text{(}|x|\le |a|\text{)}$

$\int \frac{dx}{u}=\arcsin \frac{x}{a}\text{(}|x|\le |a|\text{)}$

$\int \frac{x^{2}dx}{u}=\frac{1}{2}(-xu+a^2\arcsin \frac{x}{a})\text{(}|x|\le |a|\text{)}$

$\int udx=\frac{1}{2}(xu-\mathrm{sgn}x\mathrm{arcosh}\left|\frac{x}{a}\right|)\text{(for }|x|\ge |a|\text{)}$

$\int \frac{x}{u}dx=-u\text{(}|x|\le |a|\text{)}$

Integrals involving $R=\sqrt{a{x}^2+bx+c}$

Assume (ax² + bx + c) cannot be reduced to the following expression (px + q)² for some p and q.

$\int \frac{dx}{R}=\frac{1}{\sqrt{a}}\ln |2\sqrt{a}R+2ax+b|\text{(for }a>0\text{)}$

$\int \frac{dx}{R}=\frac{1}{\sqrt{a}}\mathrm{arsinh}\frac{2ax+b}{\sqrt{4ac-b^2}}\text{(for }a>0\text{, }4ac-b^2>0\text{)}$

$\int \frac{dx}{R}=\frac{1}{\sqrt{a}}\ln |2ax+b|\text{(for }a>0\text{, }4ac-b^2=0\text{)}$

$\int \frac{dx}{R}=-\frac{1}{\sqrt{-a}}\arcsin \frac{2ax+b}{\sqrt{b^2-4ac}}\text{(for }a<0\text{, }4ac-b^2<0\text{, }|2ax+b|<\sqrt{b^2-4ac}\text{)}$

$\int \frac{dx}{R^3}=\frac{4ax+2b}{(4ac-b^2)R}$

$\int \frac{dx}{R^5}=\frac{4ax+2b}{3(4ac-b^2)R}(\frac{1}{R^2}+\frac{8a}{4ac-b^2})$

$\int \frac{dx}{R^{2n+1}}=\frac{2}{(2n-1)(4ac-b^2)}(\frac{2ax+b}{R^{2n-1}}+4a(n-1)\int \frac{dx}{R^{2n-1}})$

$\int \frac{x}{R}dx=\frac{R}{a}-\frac{b}{2a}\int \frac{dx}{R}$

$\int \frac{x}{R^3}dx=-\frac{2bx+4c}{(4ac-b^2)R}$

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

$\int \frac{dx}{xR}=-\frac{1}{\sqrt{c}}\ln \left(\frac{2\sqrt{c}R+bx+2c}{x}\right),c>0$

$\int \frac{dx}{xR}=-\frac{1}{\sqrt{c}}\mathrm{arsinh}\left(\frac{bx+2c}{|x|\sqrt{4ac-b^2}}\right),c<0$

$\int \frac{dx}{xR}=\frac{1}{\sqrt{-c}}\arcsin \left(\frac{bx+2c}{|x|\sqrt{b^2-4ac}}\right),c<0,b^2-4ac>0$

$\int \frac{dx}{xR}=-\frac{2}{bx}\left(\sqrt{a{x}^2+bx}\right),c=0$

Integrals involving $S=\sqrt{ax+b}$

$\int Sdx=\frac{2S^3}{3a}$

$\int \frac{dx}{S}=\frac{2S}{a}$

$\int \frac{dx}{xS}=\{\begin{array}{cc}-\frac{2}{\sqrt{b}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{h}\left(\frac{S}{\sqrt{b}}\right)& \text{(for }b>0,ax>0\text{)}\\ -\frac{2}{\sqrt{b}}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{h}\left(\frac{S}{\sqrt{b}}\right)& \text{(for }b>0,ax<0\text{)}\\ \frac{2}{\sqrt{-b}}\arctan \left(\frac{S}{\sqrt{-b}}\right)& \text{(for }b<0\text{)}\end{array}$

$\int \frac{S}{x}dx=\{\begin{array}{cc}2(S-\sqrt{b}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{h}\left(\frac{S}{\sqrt{b}}\right))& \text{(for }b>0,ax>0\text{)}\\ 2(S-\sqrt{b}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{h}\left(\frac{S}{\sqrt{b}}\right))& \text{(for }b>0,ax<0\text{)}\\ 2(S-\sqrt{-b}\arctan \left(\frac{S}{\sqrt{-b}}\right))& \text{(for }b<0\text{)}\end{array}$

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

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

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