فهرست‌های انتگرال‌ها

${\displaystyle \int \frac{1}{x}dx=\ln \left|x\right|+C}$

${\displaystyle \int kdx=kx+C}$

${\displaystyle \int x^{a}dx=\frac{x^{a+1}}{a+1}+C}$

(Cavalieri's quadrature formula)

${\displaystyle \int (ax+b{)}^{n}dx=\frac{(ax+b{)}^{n+1}}{a(n+1)}+C\text{(for }n\ne -1\text{)}}$

${\displaystyle \int \frac{1}{x}dx=\ln \left|x\right|+C}$

${\displaystyle \int \frac{c}{ax+b}dx=\frac{c}{a}\ln \left|ax+b\right|+C}$

${\displaystyle \int e^{x}dx=e^x+C}$

${\displaystyle \int a^{x}dx=\frac{a^x}{\ln a}-C}$

${\displaystyle \int \ln xdx=x\ln x-x+C}$

${\displaystyle \int {\log }_{a}xdx=x{\log }_{a}x-\frac{x}{\ln a}+C}$

${\displaystyle \int \sin xdx=-\cos x+C}$

${\displaystyle \int \cos xdx=\sin x+C}$

${\displaystyle \int \tan xdx=-\ln \left|\cos x\right|+C=\ln \left|\sec x\right|+C}$

${\displaystyle \int \cot xdx=\ln \left|\sin x\right|+C}$

${\displaystyle \int \sec xdx=\ln \left|\sec x+\tan x\right|+C}$

${\displaystyle \int \csc xdx=-\ln \left|\csc x+\cot x\right|+C}$

${\displaystyle \int {\sec }^{2}xdx=\tan x+C}$

${\displaystyle \int {\csc }^{2}xdx=-\cot x+C}$

${\displaystyle \int \sec x\tan xdx=\sec x+C}$

${\displaystyle \int \csc x\cot xdx=-\csc x+C}$

${\displaystyle \int {\sin }^{2}xdx=\frac{1}{2}\left(x-\frac{\sin 2x}{2}\right)+C=\frac{1}{2}(x-\sin x\cos x)+C}$

${\displaystyle \int {\cos }^{2}xdx=\frac{1}{2}\left(x+\frac{\sin 2x}{2}\right)+C=\frac{1}{2}(x+\sin x\cos x)+C}$

${\displaystyle \int {\sec }^{3}xdx=\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln |\sec x+\tan x|+C}$

(ببینید انتگرال مکعب سکانت)

${\displaystyle \int {\sin }^{n}xdx=-\frac{{\sin }^{n-1}x\cos x}{n}+\frac{n-1}{n}\int {\sin }^{n-2}xdx}$

${\displaystyle \int {\cos }^{n}xdx=\frac{{\cos }^{n-1}x\sin x}{n}+\frac{n-1}{n}\int {\cos }^{n-2}xdx}$

${\displaystyle \int \arcsin xdx=x\arcsin x+\sqrt{1-x^2}+C,\text{ for }|x|\le +1}$

${\displaystyle \int \arccos xdx=x\arccos x-\sqrt{1-x^2}+C,\text{ for }|x|\le +1}$

${\displaystyle \int \arctan xdx=x\arctan x-\frac{1}{2}\ln |1+x^2|+C,\text{ for all real }x}$

${\displaystyle \int \mathrm{arccot}xdx=x\mathrm{arccot}x+\frac{1}{2}\ln |1+x^2|+C,\text{ for all real }x}$

${\displaystyle \int \mathrm{arcsec}xdx=x\mathrm{arcsec}x-\ln \left|x\left(1+\sqrt{1-x^{-2}}\right)\right|+C,\text{ for }|x|\ge 1}$

${\displaystyle \int \mathrm{arccsc}xdx=x\mathrm{arccsc}x+\ln \left|x\left(1+\sqrt{1-x^{-2}}\right)\right|+C,\text{ for }|x|\ge 1}$

${\displaystyle \int \sinh xdx=\cosh x+C}$

${\displaystyle \int \cosh xdx=\sinh x+C}$

${\displaystyle \int \tanh xdx=\ln (\cosh x)+C}$

${\displaystyle \int \coth xdx=\ln |\sinh x|+C,\text{ for }x\ne 0}$

${\displaystyle \int \mathrm{sech}xdx=\arctan (\sinh x)+C}$

${\displaystyle \int \mathrm{csch}xdx=\ln \left|\tanh \frac{x}{2}\right|+C,\text{ for }x\ne 0}$

${\displaystyle \int \mathrm{arsinh}xdx=x\mathrm{arsinh}x-\sqrt{x^2+1}+C,\text{ for all real }x}$

${\displaystyle \int \mathrm{arcosh}xdx=x\mathrm{arcosh}x-\sqrt{x^2-1}+C,\text{ for }x\ge 1}$

${\displaystyle \int \mathrm{arcoth}xdx=x\mathrm{arcoth}x+\frac{\ln \left(x^2-1\right)}{2}+C,\text{ for }|x|>1}$

${\displaystyle \int \mathrm{arcsch}xdx=x\mathrm{arcsch}x+|\mathrm{arsinh}x|+C,\text{ for }x\ne 0}$

${\displaystyle \int \cos ax{e}^{bx}dx=\frac{e^{bx}}{a^2+b^2}\left(a\sin ax+b\cos ax\right)+C}$

${\displaystyle \int \sin ax{e}^{bx}dx=\frac{e^{bx}}{a^2+b^2}\left(b\sin ax-a\cos ax\right)+C}$

${\displaystyle \int \cos ax\cosh bxdx=\frac{1}{a^2+b^2}\left(a\sin ax\cosh bx+b\cos ax\sinh bx\right)+C}$

${\displaystyle \int \sin ax\cosh bxdx=\frac{1}{a^2+b^2}\left(b\sin ax\sinh bx-a\cos ax\cosh bx\right)+C}$

${\displaystyle \int \left|(ax+b{)}^n\right|dx=\frac{(ax+b{)}^{n+2}}{a(n+1)\left|ax+b\right|}+C[n\text{ is odd, and }n\ne -1]}$

${\displaystyle \int \left|\sin ax\right|dx=\frac{-1}{a}\left|\sin ax\right|\cot ax+C}$

${\displaystyle \int \left|\cos ax\right|dx=\frac{1}{a}\left|\cos ax\right|\tan ax+C}$

${\displaystyle \int \left|\tan ax\right|dx=\frac{\tan (ax)[-\ln \left|\cos ax\right|]}{a\left|\tan ax\right|}+C}$

${\displaystyle \int \left|\csc ax\right|dx=\frac{-\ln \left|\csc ax+\cot ax\right|\sin ax}{a\left|\sin ax\right|}+C}$

${\displaystyle \int \left|\sec ax\right|dx=\frac{\ln \left|\sec ax+\tan ax\right|\cos ax}{a\left|\cos ax\right|}+C}$

${\displaystyle \int \left|\cot ax\right|dx=\frac{\tan (ax)[\ln \left|\sin ax\right|]}{a\left|\tan ax\right|}+C}$

Ci, Si: انتگرال مثلثاتی، Ei: انتگرال نمایی، li: انتگرال لگاریتمی، erf: تابع خطا

${\displaystyle \int \mathrm{Ci}(x)dx=x\mathrm{Ci}(x)-\sin x}$

${\displaystyle \int \mathrm{Si}(x)dx=x\mathrm{Si}(x)+\cos x}$

${\displaystyle \int \mathrm{Ei}(x)dx=x\mathrm{Ei}(x)-e^x}$

${\displaystyle \int \mathrm{li}(x)dx=x\mathrm{li}(x)-\mathrm{Ei}(2\ln x)}$

${\displaystyle \int \frac{\mathrm{li}(x)}{x}dx=\ln x\mathrm{li}(x)-x}$

${\displaystyle \int \mathrm{erf}(x)dx=\frac{e^{-x^2}}{\sqrt{\pi }}+x\text{erf}(x)}$

${\displaystyle {\int }_0^{\mathrm{\infty }}\sqrt{x}{e}^{-x}dx=\frac{1}{2}\sqrt{\pi }}$

(همچنین ببینید تابع گاما)

${\displaystyle {\int }_0^{\mathrm{\infty }}{e}^{-a{x}^2}dx=\frac{1}{2}\sqrt{\frac{\pi }{a}}}$

(انتگرال گاوسی)

${\displaystyle {\int }_0^{\mathrm{\infty }}{x}^2{e}^{-a{x}^2}dx=\frac{1}{4}\sqrt{\frac{\pi }{a^3}}}$

when a> 0

${\displaystyle {\int }_0^{\mathrm{\infty }}{x}^{2n}{e}^{-a{x}^2}dx=\frac{2n-1}{2a}{\int }_0^{\mathrm{\infty }}{x}^{2(n-1)}{e}^{-a{x}^2}dx=\frac{(2n-1)!!}{2^{n+1}}\sqrt{\frac{\pi }{a^{2n+1}}}=\frac{(2n)!}{n!2^{2n+1}}\sqrt{\frac{\pi }{a^{2n+1}}}}$

هنگامی که a> 0, n is 1,2،۳,... و !! است فاکتوریل.

${\displaystyle {\int }_0^{\mathrm{\infty }}{x}^3{e}^{-a{x}^2}dx=\frac{1}{2a^2}}$

هنگامی که a> 0

${\displaystyle {\int }_0^{\mathrm{\infty }}{x}^{2n+1}{e}^{-a{x}^2}dx=\frac{n}{a}{\int }_0^{\mathrm{\infty }}{x}^{2n-1}{e}^{-a{x}^2}dx=\frac{n!}{2a^{n+1}}}$

هنگامی که a> 0, n است ۰, ۱, ۲, ....

${\displaystyle {\int }_0^{\mathrm{\infty }}\frac{x}{e^x-1}dx=\frac{{\pi }^2}{6}}$

(همچنین ببینید Bernoulli number)

${\displaystyle {\int }_0^{\mathrm{\infty }}\frac{x^3}{e^x-1}dx=\frac{{\pi }^4}{15}}$

${\displaystyle {\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}dx=\frac{\pi }{2}}$

(see تابع سینک و انتگرال سینوسی)

${\displaystyle {\int }_0^{\mathrm{\infty }}\frac{{\sin }^{2}x}{x^2}dx=\frac{\pi }{2}}$

${\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx={\int }_0^{\frac{\pi }{2}}{\cos }^{n}xdx=\frac{1\cdot 3\cdot 5\cdot \cdots \cdot (n-1)}{2\cdot 4\cdot 6\cdot \cdots \cdot n}\frac{\pi }{2}}$

(if n is an even integer and ${\displaystyle n\ge 2}$

)

${\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx={\int }_0^{\frac{\pi }{2}}{\cos }^{n}xdx=\frac{2\cdot 4\cdot 6\cdot \cdots \cdot (n-1)}{3\cdot 5\cdot 7\cdot \cdots \cdot n}}$

(if ${\displaystyle n}$

is an odd integer and ${\displaystyle n\ge 3}$

)

(for ${\displaystyle \alpha ,\beta ,m,n}$

integers with ${\displaystyle \beta \ne 0}$

and ${\displaystyle m,n\ge 0}$

، همچنین ببینید Binomial coefficient)

${\displaystyle {\int }_{-\pi }^{\pi }\sin (\alpha x){\cos }^n(\beta x)dx=0}$

(for ${\displaystyle \alpha ,\beta }$

real and ${\displaystyle n}$

non-negative integer, همچنین ببینید تقارن)

(for ${\displaystyle \alpha ,\beta ,m,n}$

integers with ${\displaystyle \beta \ne 0}$

and ${\displaystyle m,n\ge 0}$

، همچنین ببینید Binomial coefficient)

(for ${\displaystyle \alpha ,\beta ,m,n}$

integers with ${\displaystyle \beta \ne 0}$

and ${\displaystyle m,n\ge 0}$

، همچنین ببینید Binomial coefficient)

${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-(a{x}^2+bx+c)}dx=\sqrt{\frac{\pi }{a}}\exp \left[\frac{b^2-4ac}{4a}\right]}$

(where ${\displaystyle \exp [u]}$

is the تابع نمایی ${\displaystyle e^u}$

، and ${\displaystyle a>0}$

)

${\displaystyle {\int }_0^{\mathrm{\infty }}{x}^{z-1}{e}^{-x}dx=\mathrm{\Gamma }(z)}$

(where ${\displaystyle \mathrm{\Gamma }(z)}$

is the تابع گاما)

${\displaystyle {\int }_0^1{x}^{m-1}(1-x{)}^{n-1}dx=\frac{\mathrm{\Gamma }(m)\mathrm{\Gamma }(n)}{\mathrm{\Gamma }(m+n)}}$

(the تابع بتا)

${\displaystyle {\int }_0^{2\pi }{e}^{x\cos \theta }d\theta =2\pi I_0(x)}$

(where ${\displaystyle I_0(x)}$

is the modified تابع بسل of the first kind)

${\displaystyle {\int }_0^{2\pi }{e}^{x\cos \theta +y\sin \theta }d\theta =2\pi I_0\left(\sqrt{x^2+y^2}\right)}$

${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}(1+x^2/\nu {)}^{-(\nu +1)/2}dx=\frac{\sqrt{\nu \pi }\mathrm{\Gamma }(\nu /2)}{\mathrm{\Gamma }((\nu +1)/2)}}$

، ${\displaystyle \nu >0}$

، this is related to the تابع چگالی احتمال of the توزیع تی-استیودنت)

${\displaystyle {\int }_a^{b}f(x)dx=(b-a)\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{2^n-1}{\left(-1\right)}^{m+1}{2}^{-n}f(a+m\left(b-a\right)2^{-n}).}$

${\displaystyle {\int }_0^1[\ln (1/x){]}^{p}dx=p!}$

Start by using the substitution ${\displaystyle x=\mathrm{artanh}t}$

${\displaystyle I_p={\int }_0^1[\ln (1/x){]}^p\mathrm{d}x={\int }_0^{\mathrm{\infty }}{\left[\ln (1/\mathrm{artanh}t)\right]}^p\frac{\mathrm{d}t}{1-t^2}}$

This brings the integral to the general form

${\displaystyle I_n={\int }_a^b(\ln f{)}^n{f}^{{}^{\prime }}\mathrm{d}t}$

which after integration by parts yields

${\displaystyle {\left[f(\ln f{)}^n\right]}_a^b-n{\int }_a^b(\ln f{)}^{n-1}{f}^{{}^{\prime }}\mathrm{d}t}$

and provided the first term vanishes at the end points, we get the recurrence relation

${\displaystyle I_n=-n{I}_{n-1}}$

which upon computation gives

${\displaystyle I_n=(-1{)}^{n}n!}$

Applying to our integral, we notice that

${\displaystyle [\ln (1/x){]}^p=(-1{)}^p[\ln (x){]}^p}$

Hence the final answer is:

${\displaystyle I_p=(-1{)}^p(-1{)}^{p}p!=p!}$

• M. Abramowitz and I.A. Stegun, editors. Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables.
• I.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.)
• A.P. Prudnikov (А. П. Прудников), Yu.A. Brychkov (Ю. А. Брычков), O.I. Marichev (О. И. Маричев). Integrals and Series. First edition (Russian), volume 1–5, Nauka, 1981−1986. First edition (English, translated from the Russian by N.M. Queen), volume 1–5, Gordon & Breach Science Publishers/انتشارات سی‌آرسی، 1988–1992, شابک ‎۲−۸۸۱۲۴−۰۹۷−۶. Second revised edition (Russian), volume 1–3, Fiziko-Matematicheskaya Literatura, 2003.
• Yu.A. Brychkov (Ю. А. Брычков), Handbook of Special Functions: Derivatives, Integrals, Series and Other Formulas. Russian edition, Fiziko-Matematicheskaya Literatura, 2006. English edition, Chapman & Hall/CRC Press, 2008, ISBN 1-58488-956-X.
• Daniel Zwillinger. CRC Standard Mathematical Tables and Formulae, 31st edition. Chapman & Hall/CRC Press, 2002. ISBN 1-58488-291-3. (Many earlier editions as well.)

• Meyer Hirsch, Integraltafeln, oder, Sammlung von Integralformeln (Duncker und Humblot, Berlin, 1810)
• Meyer Hirsch, Integral Tables, Or, A Collection of Integral Formulae (Baynes and son, London, 1823) [English translation of Integraltafeln]
• David Bierens de Haan, Nouvelles Tables d'Intégrales définies (Engels, Leiden, 1862)
• Benjamin O. Pierce A short table of integrals - revised edition (Ginn & co. , Boston, 1899)

• S.O.S. Mathematics: Tables and Formulas
• Paul's Online Math Notes
• A. Dieckmann, Table of Integrals (Elliptic Functions, Square Roots, Inverse Tangents and More Exotic Functions): انتگرال‌های نامعین انتگرال‌های معین
• O'Brien, Francis J. Jr. ۵۰۰ انتگرال Derived integrals of exponential and logarithmic functions
• Rule-based Mathematics Precisely defined indefinite integration rules covering a wide class of integrands

• V. H. Moll, The Integrals in Gradshteyn and Ryzhik

• Integration examples for Wolfram Alpha
• mathquick Calculator

• wxmaxima gui for Symbolic and numeric resolution of many mathematical problems