Tuesday, August 24, 2010

Formula of integrals of irrational functions

Integrals involving r = \sqrt{x^2+a^2}

\int r \;dx = \frac{1}{2}\left(x r +a^2\,\ln\left(x+r\right)\right)
\int r^3 \;dx = \frac{1}{4}xr^3+\frac{3}{8}a^2xr+\frac{3}{8}a^4\ln\left(x+r\right)
\int r^5 \; dx = \frac{1}{6}xr^5+\frac{5}{24}a^2xr^3+\frac{5}{16}a^4xr+\frac{5}{16}a^6\ln\left(x+r\right)
\int x r\;dx=\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 r\;dx= \frac{xr^3}{4}-\frac{a^2xr}{8}-\frac{a^4}{8}\ln\left(x+r\right)
\int x^2 r^3\;dx= \frac{xr^5}{6}-\frac{a^2xr^3}{24}-\frac{a^4xr}{16}-\frac{a^6}{16}\ln\left(x+r\right)
\int x^3 r \; dx = \frac{r^5}{5} - \frac{a^2 r^3}{3}
\int x^3 r^3 \; dx = \frac{r^7}{7}-\frac{a^2r^5}{5}
\int x^3 r^{2n+1} \; dx = \frac{r^{2n+5}}{2n+5} - \frac{a^3 r^{2n+3}}{2n+3}
\int x^4 r\;dx= \frac{x^3r^3}{6}-\frac{a^2xr^3}{8}+\frac{a^4xr}{16}+\frac{a^6}{16}\ln\left(x+r\right)
\int x^4 r^3\;dx= \frac{x^3r^5}{8}-\frac{a^2xr^5}{16}+\frac{a^4xr^3}{64}+\frac{3a^6xr}{128}+\frac{3a^8}{128}\ln\left(x+r\right)
\int x^5 r \; dx = \frac{r^7}{7} - \frac{2 a^2 r^5}{5} + \frac{a^4 r^3}{3}
\int x^5 r^3 \; dx = \frac{r^9}{9} - \frac{2 a^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^2r^{2n+5}}{2n+5}+\frac{a^4 r^{2n+3}}{2n+3}
\int\frac{r\;dx}{x} = r-a\ln\left|\frac{a+r}{x}\right| = r - a\, \operatorname{arsinh}\frac{a}{x}
\int\frac{r^3\;dx}{x} = \frac{r^3}{3}+a^2r-a^3\ln\left|\frac{a+r}{x}\right|
\int\frac{r^5\;dx}{x} = \frac{r^5}{5}+\frac{a^2r^3}{3}+a^4r-a^5\ln\left|\frac{a+r}{x}\right|
\int\frac{r^7\;dx}{x} = \frac{r^7}{7}+\frac{a^2r^5}{5}+\frac{a^4r^3}{3}+a^6r-a^7\ln\left|\frac{a+r}{x}\right|
\int\frac{dx}{r} = \operatorname{arsinh}\frac{x}{a} = \ln\left( \frac{x+r}{a} \right)
\int\frac{dx}{r^3} = \frac{x}{a^2r}
\int\frac{x\,dx}{r} = r
\int\frac{x\,dx}{r^3} = -\frac{1}{r}
\int\frac{x^2\;dx}{r} = \frac{x}{2}r-\frac{a^2}{2}\,\operatorname{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}\,\operatorname{arsinh}\frac{a}{x} = -\frac{1}{a}\ln\left|\frac{a+r}{x}\right|

[edit] Integrals involving s = \sqrt{x^2-a^2}

Assume (x2 > a2), for (x2 < a2), see next section:

\int s\;dx = \frac{1}{2}\left( xs-a^{2}\ln(x+s)\right)
\int xs\;dx = \frac{1}{3}s^3
\int\frac{s\;dx}{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{sgn}(x)\,\operatorname{arcosh}\left|\frac{x}{a}\right| =\frac{1}{2}\ln\left(\frac{x+s}{x-s}\right), where the positive value of \operatorname{arcosh}\left|\frac{x}{a}\right| is to be taken.

\int\frac{x\;dx}{s} = s
\int\frac{x\;dx}{s^3} = -\frac{1}{s}
\int\frac{x\;dx}{s^5} = -\frac{1}{3s^3}
\int\frac{x\;dx}{s^7} = -\frac{1}{5s^5}
\int\frac{x\;dx}{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^3s}{4}+\frac{3}{8}a^2xs+\frac{3}{8}a^4\ln\left|\frac{x+s}{a}\right|
\int\frac{x^4\;dx}{s^3} = \frac{xs}{2}-\frac{a^2x}{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|
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\int\frac{dx}{s^3}=-\frac{1}{a^2}\frac{x}{s}
\int\frac{dx}{s^5}=\frac{1}{a^4}\left[\frac{x}{s}-\frac{1}{3}\frac{x^3}{s^3}\right]
\int\frac{dx}{s^7} =-\frac{1}{a^6}\left[\frac{x}{s}-\frac{2}{3}\frac{x^3}{s^3}+\frac{1}{5}\frac{x^5}{s^5}\right]
\int\frac{dx}{s^9} =\frac{1}{a^8}\left[\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}\right]
\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}\left[\frac{1}{3}\frac{x^3}{s^3}-\frac{1}{5}\frac{x^5}{s^5}\right]
\int\frac{x^2\;dx}{s^9} = -\frac{1}{a^6}\left[\frac{1}{3}\frac{x^3}{s^3}-\frac{2}{5}\frac{x^5}{s^5}+\frac{1}{7}\frac{x^7}{s^7}\right]

[edit] Integrals involving u = \sqrt{a^2-x^2}

\int u \;dx = \frac{1}{2}\left(xu+a^2\arcsin\frac{x}{a}\right) \qquad\mbox{(}|x|\leq|a|\mbox{)}
\int xu\;dx = -\frac{1}{3} u^3 \qquad\mbox{(}|x|\leq|a|\mbox{)}
\int x^2u\;dx = -\frac{x}{4} u^3+\frac{a^2}{8}(xu+a^2\arcsin\frac{x}{a}) \qquad\mbox{(}|x|\leq|a|\mbox{)}
\int\frac{u\;dx}{x} = u-a\ln\left|\frac{a+u}{x}\right| \qquad\mbox{(}|x|\leq|a|\mbox{)}
\int\frac{dx}{u} = \arcsin\frac{x}{a} \qquad\mbox{(}|x|\leq|a|\mbox{)}
\int\frac{x^2\;dx}{u} = \frac{1}{2}\left(-xu+a^2\arcsin\frac{x}{a}\right) \qquad\mbox{(}|x|\leq|a|\mbox{)}
\int u\;dx = \frac{1}{2}\left(xu-\sgn x\,\operatorname{arcosh}\left|\frac{x}{a}\right|\right) \qquad\mbox{(for }|x|\ge|a|\mbox{)}
\int \frac{x}{u}\;dx = -u \qquad\mbox{(}|x|\leq|a|\mbox{)}

[edit] Integrals involving R = \sqrt{ax^2+bx+c}

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

0\mbox{)}" src="http://upload.wikimedia.org/math/b/9/7/b9786694b8364705405868b754f3e73a.png">
0\mbox{, }4ac-b^2>0\mbox{)}" src="http://upload.wikimedia.org/math/0/d/6/0d6ddd4644254d630d3993720801c6ad.png">
0\mbox{, }4ac-b^2=0\mbox{)}" src="http://upload.wikimedia.org/math/6/c/f/6cfab7115ce8ce8a82a1d238a1a80b7b.png">
\int\frac{dx}{R} = -\frac{1}{\sqrt{-a}}\arcsin\frac{2ax+b}{\sqrt{b^2-4ac}} \qquad \mbox{(for }a<0\mbox{, }4ac-b^2<0\mbox{, }\left|2ax+b\right|<\sqrt{b^2-4ac}\mbox{)}
\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}\left(\frac{1}{R^2}+\frac{8a}{4ac-b^2}\right)
\int\frac{dx}{R^{2n+1}} = \frac{2}{(2n-1)(4ac-b^2)}\left(\frac{2ax+b}{R^{2n-1}}+4a(n-1)\int\frac{dx}{R^{2n-1}}\right)
\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)aR^{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)
\int\frac{dx}{xR}=-\frac{1}{\sqrt{c}}\operatorname{arsinh}\left(\frac{bx+2c}{|x|\sqrt{4ac-b^2}}\right)

[edit] Integrals involving S = \sqrt{ax+b}

\int S {dx} = \frac{2 S^{3}}{3 a}
\int \frac{dx}{S} = \frac{2S}{a}
0, \quad a x > 0\mbox{)} \\ -\frac{2}{\sqrt{b}} \mathrm{artanh}\left( \frac{S}{\sqrt{b}}\right) & \mbox{(for }b > 0, \quad a x < src="http://upload.wikimedia.org/math/1/e/b/1ebd0028aa19fcf88af1c1ac87e293b9.png">
0, \quad a x > 0\mbox{)} \\ 2 \left( S - \sqrt{b}\,\mathrm{artanh}\left( \frac{S}{\sqrt{b}}\right)\right) & \mbox{(for }b > 0, \quad a x < src="http://upload.wikimedia.org/math/6/1/2/612fd00fa1ae63401b55deaa24ee002f.png">
\int \frac{x^{n}}{S} dx = \frac{2}{a (2 n + 1)} \left( x^{n} S - b n \int \frac{x^{n - 1}}{S} dx\right)
\int x^{n} S dx = \frac{2}{a (2 n + 3)} \left(x^{n} S^{3} - n b \int x^{n - 1} S dx\right)
\int \frac{1}{x^{n} S} dx = -\frac{1}{b (n - 1)} \left( \frac{S}{x^{n - 1}} + \left( n - \frac{3}{2}\right) a \int \frac{dx}{x^{n - 1} S}\right)
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