{{위키데이터 속성 추적}}
아래 목록은 [[무리 함수]]의 [[부정적분]]이다.

아래의 적분식들에서 [[적분상수]]는 생략하였다.

== ''r'' = {{sqrt|''a''<sup>2</sup> + ''x''<sup>2</sup>}}을 포함하는 함수의 적분 ==

: <math>\int r\,dx = \frac{1}{2}\left(x r +a^2\,\ln\left(x+r\right)\right)</math>

: <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>

== ''s'' = {{sqrt|''x''<sup>2</sup> − ''a''<sup>2</sup>}}을 포함하는 함수의 적분 ==
아래 적분식들에서 ''x''<sup>2</sup> > ''a''<sup>2</sup>이다. (''x''<sup>2</sup> < ''a''<sup>2</sup>인 경우는 다음 목록에 있다.)

: <math>\int s\,dx = \frac{1}{2}\left(xs-a^{2}\ln\left|x+s\right|\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>

여기서 <math>\ln\left|\frac{x+s}{a}\right|
=\operatorname{sgn}(x)\,\operatorname{arcosh}\left|\frac{x}{a}\right|
=\frac{1}{2}\ln\left(\frac{x+s}{x-s}\right)</math>이고, <math>\operatorname{arcosh}\left|\frac{x}{a}\right|</math>의 양수값을 취한다.

: <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>

== ''u'' = {{sqrt|''a''<sup>2</sup> − ''x''<sup>2</sup>}}을 포함하는 함수의 적분 ==

: <math>\int u\,dx = \frac{1}{2}\left(xu+a^2\arcsin\frac{x}{a}\right) \qquad\mbox{(}|x|\leq|a|\mbox{)}</math>

: <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>

: <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>

: <math>\int u\,dx = \frac{1}{2}\left(xu-\sgn x\,\operatorname{arcosh}\left|\frac{x}{a}\right|\right) \qquad\mbox{(}|x|\leq|a|\mbox{)}</math>

: <math>\int \frac{x}{u}\,dx = -u  \qquad\mbox{(}|x|\leq|a|\mbox{)}</math>

== ''R'' = {{sqrt|''ax''<sup>2</sup> + ''bx'' + ''c''}}를 포함하는 함수의 적분 ==

여기서 (''ax''<sup>2</sup> + ''bx'' + ''c'')는 (''px'' + ''q'')<sup>2</sup>의 꼴로 나타낼 수 없다고 가정한다.

: <math>\int\frac{dx}{R} = \frac{1}{\sqrt{a}}\ln\left|2\sqrt{a}R+2ax+b\right| \qquad \mbox{(}a>0\mbox{)}</math>

: <math>\int\frac{dx}{R} = \frac{1}{\sqrt{a}}\,\operatorname{arsinh}\frac{2ax+b}{\sqrt{4ac-b^2}} \qquad \mbox{(}a>0\mbox{, }4ac-b^2>0\mbox{)}</math>

: <math>\int\frac{dx}{R} = \frac{1}{\sqrt{a}}\ln|2ax+b| \quad \mbox{(}a>0\mbox{, }4ac-b^2=0\mbox{)}</math>

: <math>\int\frac{dx}{R} = -\frac{1}{\sqrt{-a}}\arcsin\frac{2ax+b}{\sqrt{b^2-4ac}} \qquad \mbox{(}a<0\mbox{, }4ac-b^2<0\mbox{, }\left|2ax+b\right|<\sqrt{b^2-4ac}\mbox{)}</math>

: <math>\int\frac{dx}{R^3} = \frac{4ax+2b}{(4ac-b^2)R}</math>

: <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>

: <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>

: <math>\int\frac{x}{R}\,dx = \frac{R}{a}-\frac{b}{2a}\int\frac{dx}{R}</math>

: <math>\int\frac{x}{R^3}\,dx = -\frac{2bx+4c}{(4ac-b^2)R}</math>

: <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>

: <math>\int\frac{dx}{xR} = -\frac{1}{\sqrt{c}}\ln \left|\frac{2\sqrt{c}R+bx+2c}{x}\right|\quad\mbox{(}c > 0\mbox{)}</math>

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

: <math>\int\frac{dx}{xR} = -\frac{2}{bx}\left(\sqrt{ax^2+bx}\right)\quad\mbox{(}c = 0\mbox{)}</math>

: <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}+ \frac{b}{2} \int \frac{dx}{ xR}</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>

== ''S'' = {{sqrt|''ax'' + ''b''}}를 포함하는 함수의 적분 ==

: <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}
  -\dfrac{2}{\sqrt{b}} \operatorname{arcoth}\left( \dfrac{S}{\sqrt{b}}\right) & \mbox{(}b > 0, \quad a x > 0\mbox{)} \\
  -\dfrac{2}{\sqrt{b}} \operatorname{artanh}\left( \dfrac{S}{\sqrt{b}}\right) & \mbox{(}b > 0, \quad a x < 0\mbox{)} \\
  \dfrac{2}{\sqrt{-b}} \arctan\left( \dfrac{S}{\sqrt{-b}}\right)  & \mbox{(}b < 0\mbox{)} \\
\end{cases}</math>

: <math>\int\frac{S}{x}\,dx =
\begin{cases}
 2 \left( S - \sqrt{b}\,\operatorname{arcoth}\left( \dfrac{S}{\sqrt{b}}\right)\right) & \mbox{(}b > 0, \quad a x > 0\mbox{)} \\
 2 \left( S - \sqrt{b}\,\operatorname{artanh}\left( \dfrac{S}{\sqrt{b}}\right)\right) & \mbox{(}b > 0, \quad a x < 0\mbox{)} \\
 2 \left( S - \sqrt{-b} \arctan\left( \dfrac{S}{\sqrt{-b}}\right)\right) & \mbox{(}b < 0\mbox{)} \\
\end{cases}</math>

: <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>

: <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>

: <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>

== 같이 보기 ==
• [[적분표]]

== 각주 ==

* {{서적 인용|title=A Short Table of Integrals|last=Peirce|first=Benjamin Osgood|year=1929|edition=3rd revised|publisher=Ginn and Co|location=Boston|pages=16–30|chapter=Chap. 3|orig-year=1899}}
* Milton Abramowitz and Irene A. Stegun, eds., ''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].)''
* {{서적 인용|title=Table of Integrals, Series, and Products|date=2015|edition=8|publisher=Academic Press, Inc.|language=en|orig-year=October 2014|isbn=978-0-12-384933-5|lccn=2014010276|author-first1=Izrail Solomonovich|editor-last2=Moll|translator=Scripta Technica, Inc.|editor-first2=Victor Hugo|author-last1=Gradshteyn|editor-last1=Zwillinger|editor-first1=Daniel|author-last5=Jeffrey|author-first5=Alan|author-last4=Tseytlin|author-first4=Michail Yulyevich|author-last3=Geronimus|author-first3=Yuri Veniaminovich|author-last2=Ryzhik|author-first2=Iosif Moiseevich|}}

[[분류:적분]]