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__NOTOC__ | |||
The following is a list of [[integral]]s ([[antiderivative]] functions) of [[irrational function]]s. For a complete list of integral functions, see [[lists of integrals]]. Throughout this article the [[constant of integration]] is omitted for brevity. | |||
== Integrals involving <math>r = \sqrt{x^2+a^2}</math> == | |||
: <math>\int r \;dx = \frac{1}{2}\left(x r +a^2\,\ln\left(x+r\right)\right)</math><!-- (1.1) [Abramowitz & Stegun p13 3.3.41] + verified by differentiation --> | |||
: <math>\int r^3 \;dx = \frac{1}{4}xr^3+\frac{3}{8}a^2xr+\frac{3}{8}a^4\ln\left(x+r\right)</math> | |||
: <math>\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)</math> | |||
: <math>\int x r\;dx=\frac{r^3}{3}</math> | |||
: <math>\int x r^3\;dx=\frac{r^5}{5}</math> | |||
: <math>\int x r^{2n+1}\;dx=\frac{r^{2n+3}}{2n+3} </math> | |||
: <math>\int x^2 r\;dx= \frac{xr^3}{4}-\frac{a^2xr}{8}-\frac{a^4}{8}\ln\left(x+r\right)</math> | |||
: <math>\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)</math> | |||
: <math>\int x^3 r \; dx = \frac{r^5}{5} - \frac{a^2 r^3}{3}</math> | |||
: <math>\int x^3 r^3 \; dx = \frac{r^7}{7}-\frac{a^2r^5}{5} </math> | |||
: <math>\int x^3 r^{2n+1} \; dx = \frac{r^{2n+5}}{2n+5} - \frac{a^2 r^{2n+3}}{2n+3}</math> | |||
: <math>\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)</math> | |||
: <math>\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)</math> | |||
: <math>\int x^5 r \; dx = \frac{r^7}{7} - \frac{2 a^2 r^5}{5} + \frac{a^4 r^3}{3}</math> | |||
: <math>\int x^5 r^3 \; dx = \frac{r^9}{9} - \frac{2 a^2 r^7}{7} + \frac{a^4 r^5}{5}</math> | |||
: <math>\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} </math> | |||
: <math>\int\frac{r\;dx}{x} = r-a\ln\left|\frac{a+r}{x}\right| = r - a\, \operatorname{arsinh}\frac{a}{x}</math> | |||
: <math>\int\frac{r^3\;dx}{x} = \frac{r^3}{3}+a^2r-a^3\ln\left|\frac{a+r}{x}\right|</math> | |||
: <math>\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|</math> | |||
: <math>\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|</math> | |||
: <math>\int\frac{dx}{r} = \operatorname{arsinh}\frac{x}{a} = \ln\left( \frac{x+r}{a} \right)</math> | |||
: <math>\int\frac{dx}{r^3} = \frac{x}{a^2r}</math> | |||
: <math>\int\frac{x\,dx}{r} = r</math> | |||
: <math>\int\frac{x\,dx}{r^3} = -\frac{1}{r}</math> | |||
: <math>\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)</math> | |||
: <math>\int\frac{dx}{xr} = -\frac{1}{a}\,\operatorname{arsinh}\frac{a}{x} = -\frac{1}{a}\ln\left|\frac{a+r}{x}\right|</math> | |||
== Integrals involving <math>s = \sqrt{x^2-a^2}</math>== | |||
Assume <math>(x^2>a^2)</math>, for <math>(x^2<a^2)</math>, see next section: | |||
: <math>\int s\;dx = \frac{1}{2}\left( xs-a^{2}\ln(x+s)\right)</math> | |||
: <math>\int xs\;dx = \frac{1}{3}s^3</math> | |||
: <math>\int\frac{s\;dx}{x} = s - a\arccos\left|\frac{a}{x}\right|</math> | |||
: <math>\int\frac{dx}{s} = \ln\left|\frac{x+s}{a}\right|</math> | |||
Here <math>\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)</math>, where the positive value of <math>\operatorname{arcosh}\left|\frac{x}{a}\right|</math> is to be taken. | |||
: <math>\int\frac{x\;dx}{s} = s</math> | |||
: <math>\int\frac{x\;dx}{s^3} = -\frac{1}{s}</math> | |||
: <math>\int\frac{x\;dx}{s^5} = -\frac{1}{3s^3}</math> | |||
: <math>\int\frac{x\;dx}{s^7} = -\frac{1}{5s^5}</math> | |||
: <math>\int\frac{x\;dx}{s^{2n+1}} = -\frac{1}{(2n-1)s^{2n-1}} </math> | |||
: <math>\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}} | |||
</math> | |||
: <math>\int\frac{x^2\;dx}{s} | |||
= \frac{xs}{2}+\frac{a^2}{2}\ln\left|\frac{x+s}{a}\right|</math> | |||
: <math>\int\frac{x^2\;dx}{s^3} | |||
= -\frac{x}{s}+\ln\left|\frac{x+s}{a}\right|</math> | |||
: <math>\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| </math> | |||
: <math>\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| </math> | |||
: <math>\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| </math> | |||
: <math>\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}{n-m-1 \choose i}\frac{x^{2(m+i)+1}}{s^{2(m+i)+1}}\qquad\mbox{(}n>m\ge0\mbox{)}</math> | |||
: <math>\int\frac{dx}{s^3}=-\frac{1}{a^2}\frac{x}{s}</math> | |||
: <math>\int\frac{dx}{s^5}=\frac{1}{a^4}\left[\frac{x}{s}-\frac{1}{3}\frac{x^3}{s^3}\right]</math> | |||
: <math>\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]</math> | |||
: <math>\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]</math> | |||
: <math>\int\frac{x^2\;dx}{s^5}=-\frac{1}{a^2}\frac{x^3}{3s^3}</math> | |||
: <math>\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]</math> | |||
: <math>\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]</math> | |||
== Integrals involving <math>u = \sqrt{a^2-x^2}</math>== | |||
: <math>\int u \;dx = \frac{1}{2}\left(xu+a^2\arcsin\frac{x}{a}\right) \qquad\mbox{(}|x|\leq|a|\mbox{)}</math><!-- (3.1) [Abramowitz & Stegun p13 3.3.45] --> | |||
: <math>\int xu\;dx = -\frac{1}{3} u^3 \qquad\mbox{(}|x|\leq|a|\mbox{)}</math> | |||
: <math>\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{)}</math> | |||
: <math>\int\frac{u\;dx}{x} = u-a\ln\left|\frac{a+u}{x}\right| \qquad\mbox{(}|x|\leq|a|\mbox{)}</math> | |||
: <math>\int\frac{dx}{u} = \arcsin\frac{x}{a} \qquad\mbox{(}|x|\leq|a|\mbox{)}</math><!-- (3.4) [Abramowitz & Stegun p13 3.3.44] --> | |||
: <math>\int\frac{x^2\;dx}{u} = \frac{1}{2}\left(-xu+a^2\arcsin\frac{x}{a}\right) \qquad\mbox{(}|x|\leq|a|\mbox{)}</math><!-- (3.5) [need reference] - verified by differentiation only --> | |||
: <math>\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{)}</math> | |||
:<math>\int \frac{x}{u}\;dx = -u \qquad\mbox{(}|x|\leq|a|\mbox{)}</math> | |||
== Integrals involving <math>R = \sqrt{ax^2+bx+c}</math>== | |||
Assume (''ax''<sup>2</sup> + ''bx'' + ''c'') cannot be reduced to the following expression (''px'' + ''q'')<sup>2</sup> for some ''p'' and ''q''. | |||
: <math>\int\frac{dx}{R} = \frac{1}{\sqrt{a}}\ln\left|2\sqrt{a}R+2ax+b\right| \qquad \mbox{(for }a>0\mbox{)}</math><!-- (4.1) [Abramowitz & Stegun p13 3.3.33] + verified by differentiation --> | |||
: <math>\int\frac{dx}{R} = \frac{1}{\sqrt{a}}\,\operatorname{arsinh}\frac{2ax+b}{\sqrt{4ac-b^2}} \qquad \mbox{(for }a>0\mbox{, }4ac-b^2>0\mbox{)}</math><!-- (4.2) [Abramowitz & Stegun p13 3.3.34] + verified by differentiation --> | |||
: <math>\int\frac{dx}{R} = \frac{1}{\sqrt{a}}\ln|2ax+b| \quad \mbox{(for }a>0\mbox{, }4ac-b^2=0\mbox{)}</math><!-- (4.3) [Abramowitz & Stegun p13 3.3.35] + verified by differentiation --> | |||
: <math>\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{)}</math><!-- (4.4) [Abramowitz & Stegun p13 3.3.36] + verified by differentiation --> | |||
: <math>\int\frac{dx}{R^3} = \frac{4ax+2b}{(4ac-b^2)R}</math><!-- (4.5) [need reference] - verified by differentiation + special case of 4.7 below--> | |||
: <math>\int\frac{dx}{R^5} = \frac{4ax+2b}{3(4ac-b^2)R}\left(\frac{1}{R^2}+\frac{8a}{4ac-b^2}\right)</math><!-- (4.6) [need reference] - verified by differentiation + special case of 4.7 below--> | |||
: <math>\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)</math><!-- (4.7) [need reference] - verified by differentiation only --> | |||
: <math>\int\frac{x}{R}\;dx = \frac{R}{a}-\frac{b}{2a}\int\frac{dx}{R}</math><!-- (4.8) [Abramowitz & Stegun p13 3.3.39] + verified by differentiation --> | |||
: <math>\int\frac{x}{R^3}\;dx = -\frac{2bx+4c}{(4ac-b^2)R}</math><!-- (4.9) [need reference] - verified by differentiation only --> | |||
: <math>\int\frac{x}{R^{2n+1}}\;dx = -\frac{1}{(2n-1)aR^{2n-1}}-\frac{b}{2a}\int\frac{dx}{R^{2n+1}}</math><!-- (4.10) [need reference] - verified by differentiation only --> | |||
: <math>\int\frac{dx}{xR}=-\frac{1}{\sqrt{c}}\ln \left|\frac{2\sqrt{c}R+bx+2c}{x}\right|, ~ c > 0</math><!-- (4.11) [Abramowitz & Stegun p13 implied by 3.3.38 + 3.3.33] + verified by differentiation --> | |||
: <math>\int\frac{dx}{xR}=-\frac{1}{\sqrt{c}}\operatorname{arsinh}\left(\frac{bx+2c}{|x|\sqrt{4ac-b^2}}\right), ~ c < 0</math><!-- (4.11) [Abramowitz & Stegun p13 implied by 3.3.38 + 3.3.34] + verified by differentiation --> | |||
: <math>\int\frac{dx}{xR}=\frac{1}{\sqrt{-c}}\operatorname{arcsin}\left(\frac{bx+2c}{|x|\sqrt{b^2-4ac}}\right), ~ c < 0, b^2-4ac>0</math><!-- (4.11) [Abramowitz & Stegun p13 implied by 3.3.38 + 3.3.34] + verified by differentiation --> | |||
: <math>\int\frac{dx}{xR}=-\frac{2}{bx}\left(\sqrt{ax^2+bx}\right), ~ c = 0</math> | |||
<blockquote> | |||
{{hidden | |||
|(Click "show" at right to see a proof or "hide" to hide it.) | |||
|2= | |||
---- | |||
First use substitution with <math>u \equiv \sqrt{ax+b}</math> so that <math>R = \sqrt{x} u</math>, and the integral becomes | |||
: <math>2\sqrt{a} \int \frac{du}{(u^2-b)^{\frac{3}{2}}}</math> | |||
Then make another substitution with <math>v \equiv \frac{1}{\sqrt{u^2-b}}</math> so that the integral becomes | |||
: <math> -2\sqrt{a} \int \frac{vdv}{\sqrt{1+bv^2}}</math> | |||
The integral then has the form <math>\int \frac{x\;dx}{r} = r</math> as given above and can be readily evaluated. | |||
One can also perform the substitutions in a single step with the less apparent choice <math>u \equiv \sqrt{\frac{ax + b}{ax}}</math>, giving | |||
: <math>\int\frac{dx}{xR}= -\frac{2\sqrt{a}}{b} \int du = -\frac{2\sqrt{a}}{b}u</math> | |||
---- | |||
}} | |||
</blockquote> | |||
: <math>\int\frac{x^2}{R}\;dx = \frac{2ax-3b}{4a^2}R+\frac{3b^2-4ac}{8a^2}\int\frac{dx}{R}</math> | |||
: <math> \int \frac{dx}{x^{2} R}=- \frac{ R}{cx}-\frac{b}{2c} \int \frac{dx}{x R}</math> | |||
: <math>\int R\,dx= \frac{2ax+b}{4a} R+ \frac{4ac-b^{2}}{8a} \int \frac{dx}{ R}</math> | |||
: <math>\int x R\,dx = \frac{R^3}{3a}-\frac{b(2ax+b)}{8a^{2}} R - \frac{b(4ac-b^{2})}{16a^{2}} \int \frac{dx}{ R}</math> | |||
: <math>\int x^{2} R\,dx= \frac{6ax-5b}{24a^{2}}R^3+\frac{5b^{2}-4ac}{16a^{2}} \int R\,dx</math> | |||
: <math>\int \frac{ R}{x}\,dx= R+ \frac{b}{2} \int \frac{dx}{ R}+c \int \frac{dx}{x R}</math> | |||
: <math>\int \frac{ R}{x^{2}}\,dx=- \frac{ R}{x}+a \int \frac{dx}{R^2}+ \frac{b}{2} \int \frac{dx}{ R}</math> | |||
: <math>\int \frac{x^{2}\,dx}{R^3}= \frac{(2b^{2}-4ac)x+2bc}{a(4ac-b^{2}) R}+ \frac{1}{a} \int \frac{dx}{ R}</math> | |||
== Integrals involving <math>S = \sqrt{ax+b}</math>== | |||
: <math>\int S {dx} = \frac{2 S^{3}}{3 a}</math> | |||
: <math>\int \frac{dx}{S} = \frac{2S}{a}</math> | |||
: <math>\int \frac{dx}{x S} = | |||
\begin{cases} | |||
-\frac{2}{\sqrt{b}} \mathrm{arcoth}\left( \frac{S}{\sqrt{b}}\right) & \mbox{(for }b > 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 < 0\mbox{)} \\ | |||
\frac{2}{\sqrt{-b}} \arctan\left( \frac{S}{\sqrt{-b}}\right) & \mbox{(for }b < 0\mbox{)} \\ | |||
\end{cases}</math> | |||
: <math>\int\frac{S}{x}\,dx = | |||
\begin{cases} | |||
2 \left( S - \sqrt{b}\,\mathrm{arcoth}\left( \frac{S}{\sqrt{b}}\right)\right) & \mbox{(for }b > 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 < 0\mbox{)} \\ | |||
2 \left( S - \sqrt{-b} \arctan\left( \frac{S}{\sqrt{-b}}\right)\right) & \mbox{(for }b < 0\mbox{)} \\ | |||
\end{cases}</math><!--yes it is S minus etc. in both cases as --> | |||
: <math>\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)</math><!-- (5.4) [need reference] - verified by differentiation only --> | |||
: <math>\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)</math><!-- (5.5) [need reference] - verified by differentiation only --> | |||
: <math>\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)</math> | |||
== References == | |||
* {{cite book | |||
|last=Peirce | |||
|first=Benjamin Osgood | |||
|title=A Short Table of Integrals | |||
|origyear=1899 | |||
|edition=3rd revised | |||
|year=1929 | |||
|publisher=Ginn and Co | |||
|location=Boston | |||
|pages=16–30 | |||
|chapter=Chap. 3 | |||
}} | |||
* Milton Abramowitz and Irene A. Stegun, eds., ''[[Abramowitz and Stegun|Handbook of Mathematical Functions]] with Formulas, Graphs, and Mathematical Tables'' 1972, Dover: New York. ''(See [http://www.math.sfu.ca/~cbm/aands/page_13.htm chapter 3].)'' | |||
* S. Gradshteyn (И.С. Градштейн), I.M. Ryzhik (И.М. Рыжик); Alan Jeffrey, Daniel Zwillinger, editors. Table of Integrals, Series, and Products, seventh edition. Academic Press, 2007. ISBN 978-0-12-373637-6. Errata. (Several previous editions as well.) | |||
{{Lists of integrals}} | |||
[[Category:Integrals|Irrational functions]] | |||
[[Category:Mathematics-related lists|Integrals of irrational functions]] |
Revision as of 08:45, 27 January 2014
The following is a list of integrals (antiderivative functions) of irrational functions. For a complete list of integral functions, see lists of integrals. Throughout this article the constant of integration is omitted for brevity.
Integrals involving
Integrals involving
Assume , for , see next section:
Here , where the positive value of is to be taken.
Integrals involving
Integrals involving
Assume (ax2 + bx + c) cannot be reduced to the following expression (px + q)2 for some p and q.
Integrals involving
References
- 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
My blog: http://www.primaboinca.com/view_profile.php?userid=5889534 - Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables 1972, Dover: New York. (See chapter 3.)
- S. Gradshteyn (И.С. Градштейн), I.M. Ryzhik (И.М. Рыжик); Alan Jeffrey, Daniel Zwillinger, editors. Table of Integrals, Series, and Products, seventh edition. Academic Press, 2007. ISBN 978-0-12-373637-6. Errata. (Several previous editions as well.)