비합리적인 기능의 통합 목록

다음은 불합리한 기능의 통합(반복적 기능) 목록이다. 통합 기능의 전체 목록은 통합 목록을 참조하십시오. 이 글에서 통합의 상수는 간결성을 위해 생략한다.

r = √a² + x와² 관련된 통합

${\dplaystyle \int r\,dx={\frac {1}{1}:{2}}\왼쪽(xr+a^{2}\,\ln \좌측(x+r\우측)\오른쪽)}$

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

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

${\displaystyle \int xr\,dx={\frac {r^{3}}{3}}}$

${\displaystyle \int xr^{3}\,dx={\frac{r^{5}}{5}}}$

${\displaystyle \int xr^{2n+1}\,dx={\frac {r^{2n+3}}{2n+3}}}$

${\displaystyle \int x^{2}r\,dx={\frac {xr^{3}}{4}{8}-{\frac {a^{2}xr}{8}}{8}}}\ln \좌(x+r\오른쪽)}$

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

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

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

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

${\displaystyle \int x^{4}r\,dx={\frac {x^{3}r^{6}{6}-{6}-{\frac {a^{2}xr^{3}}}}{8}}}}+{a^{4}xr}}}}{16}}}{16}}}}}}}{16}}}}}}}}}}}}}{16}}}}}}}}}}}}}}}}}{16}}}}}}}}}}}}}}}$

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

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

${\displaystyle \int x^{5}r^{3}\,dx={\frac {r^{9}}{9}}}-{\frac {2a^{2}r^{7}}}}{7}}}}}{7}}}}}}{5}}}}}}}}}}}}}}}}}}}}}}}}}}}}"$

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

${\displaystyle \int{\frac {r\,dx}{x}}=r-a\ln \왼쪽 {\frac {a+r}{x}\right =r-a\,\parsinname {arsinh} {\frac {a}{x}}}}}}}}$

${\displaystyle \int{\r^{3}\,dx}{x}}}{x}}}={\frac {r^{3}}}}+a^{3}}}}{2}-a^{3}\ln \좌측 {\frac {a+r}{x}}\오른쪽 }}$

${\displaystyle \int{\r^{5}\,dx}{x}}}{x}}}={\frac {r^{5}}+{a^{2}r^{3}}}{3}}}}+a^{4}r-a^{5}}}}}}}}{x\rig}}}}}}}}}}}}오른쪽 }}}}}}}}$

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

${\displaystyle \int{\frac {dx}{r}=\frac {x}}}{a}=\ln \frac\fract\{x+r}{a}\right)}$

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

${\displaystyle \int{\frac {x\,dx}{r}=r}$

${\displaystyle \int{\frac {x\,dx}{r^{3}=-{\frac {1}{r}}}$

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

${\displaystyle \int{\frac{dx}}{xr}=-{\frac {1}{a}\cH}{\frac {a}}=-{\frac{1}}}\ln \{\frac {a+r}{x}}}\right}}$

s = √x² - a를² 포함하는 통합

x² > a라고² 가정한다(x² < a의² 경우, 다음 섹션 참조):

${\dplaystyle \ints\,dx={\frac {1}{1}:{2}}\왼쪽(xs-a^{2}\ln \왼쪽 x+s\오른쪽 \오른쪽)}$

${\displaystyle \int xs\,dx={\frac {1}{3}s^{3}}}$

${\displaystyle \int{\frac {s\,dx}{x}}=s- \arccos \왼쪽 {\frac {a}{x}}\오른쪽 }$

${\displaystyle \int{\frac {dx}{s}}=\ln \left {\frac {x+s}{a}\오른쪽 }$

Here ${\displaystyle \ln \left {\frac {x+s}{a}}\right =\operatorname {sgn} (x)\,\operatorname {arcosh} \left {\frac {x}{a}}\right ={\frac {1}{2}}\ln \left({\frac {x+s}{x-s}}\right)}$, where the positive value of ${\displaystyle \mathrm{arcosh}\frac{x}{a}}$$\displaystyle \propertyname {oshosh} \왼쪽 {\frac {x}{a}\오른쪽 }$을(를) 찍는다$$.

${\displaystyle \int{\frac {x\,dx}{s}=s}$

${\displaystyle \int{\frac {x\,dx}{s^{3}}=-{\frac {1}{s}}}}$

${\displaystyle \int{\frac {x\,dx}{s^{5}}=-{\frac {1}{3s^{3}}}}}}}}}$

${\displaystyle \int{\frac {x\,dx}{s^{7}}=-{\frac {1}{5s^{5}}}}}}}}$

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

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

${\displaystyle \int{\x^{2}\,dx}{s}}}{s}}}={\frac {xs}{2}}+{\frac {a^{2}}}\ln \왼쪽 {\frac {x+s}{a}}}\오른쪽 }}$

${\displaystyle \int{\frac {x^{2}\,dx}{s^{3}}=-{\frac {x}}}}}-{\frac}}\ln \왼쪽 {\frac {x+s}{a}}}\오른쪽 }$

${\displaystyle \int{\frac {x^{4}\,dx}{s}}}{s}{4}}{4}}{4}}{8}a^{8}}}}}}{{3}}}}}}{8}}}}ln \prac {x+s}{a}}}}}}}오른쪽 }$

${\displaystyle \int{\x^{4}\,dx}{s^{3}}}={\frac {xs}{2}}-{a^{2}x}{s}}}}+{\frac {3}{2}}a^{n}}\{\frac {x+s}{a}}}}}}}}}}}}}}}}오른쪽 }}}}}}}}$

${\displaystyle \int{\frac {x^{4}\,dx}{s^{5}}=-{\frac {x}{1}{1}{3}}{\x^{3}}}}}}{\frac {x+s}}}}}}}{\frac {a}}}오른쪽 }}$

${\displaystyle \int {\frac {x^{2m}\,dx}{s^{2n+1}}}=(-1)^{n-m}{\frac {1}{a^{2(n-m)}}}\sum _{i=0}^{n-m-1}{\frac {1}{2(m+i)+1}}{n-m-1 \choose i}{\frac {x^{2(m+i)+1}}{s^{2(m+i)+1}}}\qquad {\mbox{(}}n>m\geq 0{\mbox{)}}}$

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

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

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

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

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

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

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

u = √a² - x와² 관련된 통합

${\dx={\displaystyle \int u\,dx}{\frac {1}{1}:{2}}\좌측(xu+a^{2)}}\arcsin {\frac {x}{a}\right)\qquad {\mbox{{}}* x \leq a {\mbox{}}}}$

${\dplaystyle \int su\,dx=-{\frac {1}{3}u^{3}\qquad {\mbox{}}* x \leq a {\mbox{}}}}$

${\displaystyle \int x^{2}u\,dx=-{\frac {x}{4}u^{3}+{\frac{a^{8}}}{8}}}(으)+a^{2}}}\arcsin {\frac {x}{a})\qquad {\mbox{{}}}* x \leq a {\mbox{}}}}$

${\displaystyle \int{\frac {u\,dx}{x}=u-a\ln \왼쪽 {\frac {a+u}{x}}\오른쪽 \qquad {\mbox{}}}* x \leq a {\mbox{\}}}}$

${\displaystyle \int{\frac {dx}{u}}=\arcsin {\frac {x}}{a}\qquad {\mbox{}}}}* x \leq a {\mbox{}}}}$

${\displaystyle \int{\frac {x^{2}\,dx}{u}}={\frac {1}{1}:{2}}\좌측 (-+a^{2}}}\arcsin {\frac {x}{a}\right)\qquad {\mbox{{}}* x \leq a {\mbox{}}}}$

${\displaystyle \int u\,dx={\frac {1}{1}{2}}\좌측(이름: {sgn}x}x}{a}\우측 \오른쪽)\{\qquad{\mbox}}}}$

${\displaystyle \int{\frac {x}{u}}\,dx=-u\qquad {\mbox{}}* x \leq a {\mbox{}}}}$

R = √ax² + bx + c와 관련된 통합

일부 p와 q에 대해 (ax² + bx + c)를 다음 식(px + q)²으로 줄일 수 없다고 가정한다.

${\displaystyle \int {\frac {dx}{R}}={\frac{1}{\sqrt{a}}\ln \왼쪽 2{\\sqrt{a}R+2ax+b\오른쪽 \qquad{{}a>0{\mbox{}}}}$

${\displaystyle \int {\frac {dx}{R}={}={\frac {1}{\sqrt{a}}\operatorname {arsinh}{\frac+b}{\sqrt{4ax-b^{2}}:}\qquad{\mbox{{}}}}}}}}{}a>0{}}}}}}}}{}}}}{}mbox{{{{{{{{{}mbox{{{}}}}}}}}}}}}}}}}}}}}})}}}$

${\displaystyle \int {\frac {dx}{R}}={\frac{1}{\sqrt{a}}\ln 2ax+b \quad{{\mbox{{}}, }4ac-b^{2}=0\mbox{})}}}$

${\displaystyle \int {\frac {dx}{R}}=-{\frac {1}{\sqrt {-a}}}\arcsin {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}\qquad {\mbox{(for }}a<0{\mbox{, }}4ac-b^{2}<0{\mbox{, }}\left 2ax+b\right <{\sqrt {b^{2}-4ac}}{\mbox{)}}}$

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

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

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

${\displaystyle \int{\frac {x}{R}\,dx={\frac {R}{a}-{\frac {b}}{2a}}\int {\frac {d}{dx}}}{\frac {dx}}}{\px}}}}}}}}}R}}$

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

${\dx}{R^{2n+1}}\,dx=-{\frac {1}{{{n-1){{{1}{{{n-1}{{{n-1}-{\frac {b}-{dx}}}}}-{{R^{dx}{n+1}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}$

${\dplaystyle \int{\frac {dx}{xR}=-{\frac {1}{\sqrt{c}}\ln \좌측 {\frac {2}{\c}R+bx+2c}{x}\오른쪽,~~c>0}$

${\displaystyle \int{\frac {dx}{xR}=-{\frac {1}{\sqrt {c}\operatorname {arsinh} \좌측({\frac {bx+2c}{x}{4ac-b^{2}}}}}}}\c<0})$

${\displaystyle \int{\frac {dx}{xR}={\frac {1}{\frac{-c}}\operatorname {arcsin} \좌측({\frac {bx+2c}{\xqrt {b^{2}-4ac}}}}}}}}\오른쪽)~c<0,b^2}$

${\dplaystyle \int{\frac {dx}{xR}=-{\frac {2}{bx}\왼쪽({\sqrt {ax^{2}+bx}\오른쪽),~~c=0}$

${\displaystyle \int{\frac{x^{2}}:{{R}\x={\frac-3b}{4a^{2}}:R+{3b^{2}-4ac}{8a^{dx}}}}}{\frac-{dx}}}{\frac}}}}{\frac {dx}}{R}}$

${\displaystyle \int {\frac {dx}{x^{2}}}}=-{\frac {R}{cx}-{\frac {b}{2c}\int{\frac {dx}{xR}}}}}}$

${\dplaystyle \int R\,dx={\frac {2ax+b}{4a}R+{\frac {4ac-b^{2}}:{8a}\frac {dx}{R}}$

${\displaystyle \int xR\,dx={\frac{3a}-{3a}-{\frac {b(2ax+b)}{8a^{2}}:R-{\b(4ac-b^{2}}}}}{16a^{2}}:}\frac {dx}{dx}{}}}}}R}}$

${\displaystyle \int x^{2}R\,dx={\frac {6ax-5b}{24a^{2}}:R^{3}+{\frac {5b^{2}-4ac}{16a^{2}}:R\int R\,dx}$

${\displaystyle \int{\frac {R}{x}\,dx=R+{\frac {b}{2}}\int {\frac {dx}{R}}+c\int {\frac {dx}{xR}}}$

${\displaystyle \int{\frac {R}{x^{2}}\\,dx=-{\frac {R}{x}}+a\\frac {dx}{dx}{{dx}}}}}}{\x}}}}}}}R}+{\frac {b}{2}}:\int {\frac {dx}{xR}}}}$

${\displaystyle \int{\frac {x^{2}\,dx}{R^{3}}={\frac {(2b^{2}-4ac)x+2bc}{a(4ac-b^{2})R}+{\frac{1}{a}\int{\frac {dx}{R}}$

S = √ax + b와 관련된 통합

${\dplaystyle \int S\,dx={\frac {2S^{3}{3a}}}$

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

${\displaystyle \int {\frac {dx}{x}S}}={\begin}-{\dfrac {2}{\sqrt{b}}\operatorname {arcot} \left({\dfrac {S}{\sqrt{b}}\오른쪽) \\\mbox{{}}}}}}}}{{}b>0,\mbox{\mbox}}}\\\-{\dfrac {2}{\sqrt {b}\operatorname {artanh} \좌측({\dfrac {S}{\sqrt {b}}\우측)&{{}\mbox{{}}}}{}}}}}}{}\mbox{{{{}}}}}}}}}}}}}}}}}}}}}}}}\mbox}}}}}}}}}}}}}}}}}}}}}}}}}}}}\\{\dfrac {2}{\sqrt {-b}}\arctan \left({\dfrac {S}{\sqrt{-b}\right)&{\mbox{}{}}}}}}{}b<0}\mbox{{})}}\\\end{case}}$

${\displaystyle \int {\frac {S}{x}}\,dx={\begin{cases}2\left(S-{\sqrt {b}}\,\operatorname {arcoth} \left({\dfrac {S}{\sqrt {b}}}\right)\right)&{\mbox{(for }}b>0,\quad ax>0{\mbox{)}}\\2\왼쪽(S-{\sqrt {b}\,\operatorname {artanh} \left({\dfrac {S}{\sqrt {b}\오른쪽)&{\mbox{}}{}}}}{}b}0,\quad ax<0{\mbox{})}}\\2\왼쪽(S-{\sqrt {-b}\arctan \left({\dfrac {S}{\sqrt {-b}\오른쪽)&{\mbox{}{}}}}{}b<0}\mbox{})}}\\\end{case}}$

${\displaystyle \int{\frac{x^{n}}{S}\,dx={n2}{a(2n+1)}}}}}\좌측(x^{n}S-bn\int {\x^{n-1}}{dx\right)}$

${\displaystyle \int x^{n}S\,dx={\frac {2}{a(2n+3)}}}\왼쪽(x^{n^{n}S^{3}-nb\int x^{n-1}S\,dx\right)}}$

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

참조

• Peirce, Benjamin Osgood (1929) [1899]. "Chap. 3". A Short Table of Integrals (3rd revised ed.). Boston: Ginn and Co. pp. 16–30.
• 밀턴 아브라모위츠와 아이린 A. Stegun, eds, Handbook of Mathemical Functions with 공식, 그래프 및 Mathemical Tables 1972, Dover: 뉴욕 (제3장 참조)
• Gradshteyn, Izrail Solomonovich, Ryzhik, 이오시프 Moiseevich, Geronimus, 유리 Veniaminovich, Tseytlin, Michail Yulyevich, 제프리, 앨런(2015년)[10월 2014년].Zwillinger, 다니엘;몰, 빅터 휴고는(eds.).Integrals, 시리즈, 그리고 상품의 테이블입니다.Scripta 테크니카, Inc.(8판).학술 출판부, Inc.아이 에스비엔 978-0-12-384933-5. LCCN 2014010276.(몇몇 이전 버전이기도 하다.).