Table of Integrals ā« u ā d v = u v ā ā« v ā d u \int u\,dv=uv-\int v\,du ā« u d v = uv ā ā« v d u ā« u n ā d u = u n + 1 n + 1 + C , n ā ā 1 \int u^n\,du=\frac{u^{n+1}}{n+1}+C,\quad n\ne-1 ā« u n d u = n + 1 u n + 1 ā + C , n ī = ā 1 ā« d u u = ln ā” ā£ u ā£ + C \int \frac{du}{u}=\ln|u|+C ā« u d u ā = ln ā£ u ā£ + C ā« e u ā d u = e u + C \int e^u\,du=e^u+C ā« e u d u = e u + C ā« b u ā d u = b u ln ā” b + C \int b^u\,du=\frac{b^u}{\ln b}+C ā« b u d u = l n b b u ā + C ā« sin ā” u ā d u = ā cos ā” u + C \int \sin u\,du=-\cos u+C ā« sin u d u = ā cos u + C ā« cos ā” u ā d u = sin ā” u + C \int \cos u\,du=\sin u+C ā« cos u d u = sin u + C ā« sec ā” 2 u ā d u = tan ā” u + C \int \sec^2u\,du=\tan u+C ā« sec 2 u d u = tan u + C ā« csc ā” 2 u ā d u = ā cot ā” u + C \int \csc^2u\,du=-\cot u+C ā« csc 2 u d u = ā cot u + C ā« sec ā” u tan ā” u ā d u = sec ā” u + C \int \sec u\tan u\,du=\sec u+C ā« sec u tan u d u = sec u + C ā« csc ā” u cot ā” u ā d u = ā csc ā” u + C \int \csc u\cot u\,du=-\csc u+C ā« csc u cot u d u = ā csc u + C ā« tan ā” u ā d u = ln ā” ā£ sec ā” u ā£ + C \int \tan u\,du=\ln|\sec u|+C ā« tan u d u = ln ā£ sec u ā£ + C ā« cot ā” u ā d u = ln ā” ā£ sin ā” u ā£ + C \int \cot u\,du=\ln|\sin u|+C ā« cot u d u = ln ā£ sin u ā£ + C ā« sec ā” u ā d u = ln ā” ā£ sec ā” u + tan ā” u ā£ + C \int \sec u\,du=\ln|\sec u+\tan u|+C ā« sec u d u = ln ā£ sec u + tan u ā£ + C ā« csc ā” u ā d u = ln ā” ā£ csc ā” u ā cot ā” u ā£ + C \int \csc u\,du=\ln|\csc u-\cot u|+C ā« csc u d u = ln ā£ csc u ā cot u ā£ + C ā« d u a 2 ā u 2 = sin ā” ā 1 u a + C , a > 0 \int \frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\frac{u}{a}+C,\quad a>0 ā« a 2 ā u 2 ā d u ā = sin ā 1 a u ā + C , a > 0 ā« d u a 2 + u 2 = 1 a tan ā” ā 1 u a + C \int \frac{du}{a^2+u^2}=\frac1a\tan^{-1}\frac{u}{a}+C ā« a 2 + u 2 d u ā = a 1 ā tan ā 1 a u ā + C ā« d u u u 2 ā a 2 = 1 a sec ā” ā 1 ā£ u a ā£ + C \int \frac{du}{u\sqrt{u^2-a^2}}=\frac1a\sec^{-1}\left|\frac{u}{a}\right|+C ā« u u 2 ā a 2 ā d u ā = a 1 ā sec ā 1 ā a u ā ā + C ā« d u a 2 ā u 2 = 1 2 a ln ā” ā£ u + a u ā a ā£ + C \int \frac{du}{a^2-u^2}=\frac1{2a}\ln\left|\frac{u+a}{u-a}\right|+C ā« a 2 ā u 2 d u ā = 2 a 1 ā ln ā u ā a u + a ā ā + C ā« d u u 2 ā a 2 = 1 2 a ln ā” ā£ u ā a u + a ā£ + C \int \frac{du}{u^2-a^2}=\frac1{2a}\ln\left|\frac{u-a}{u+a}\right|+C ā« u 2 ā a 2 d u ā = 2 a 1 ā ln ā u + a u ā a ā ā + C ā« a 2 + u 2 ā d u = u 2 a 2 + u 2 + a 2 2 ln ā” ( u + a 2 + u 2 ) + C \int\sqrt{a^2+u^2}\,du=\frac u2\sqrt{a^2+u^2}+\frac{a^2}2\ln\left(u+\sqrt{a^2+u^2}\right)+C ā« a 2 + u 2 ā d u = 2 u ā a 2 + u 2 ā + 2 a 2 ā ln ( u + a 2 + u 2 ā ) + C ā« u 2 a 2 + u 2 ā d u = u 8 ( a 2 + 2 u 2 ) a 2 + u 2 ā a 4 8 ln ā” ( u + a 2 + u 2 ) + C \int u^2\sqrt{a^2+u^2}\,du=\frac u8(a^2+2u^2)\sqrt{a^2+u^2}-\frac{a^4}8\ln\left(u+\sqrt{a^2+u^2}\right)+C ā« u 2 a 2 + u 2 ā d u = 8 u ā ( a 2 + 2 u 2 ) a 2 + u 2 ā ā 8 a 4 ā ln ( u + a 2 + u 2 ā ) + C ā« a 2 + u 2 u ā d u = a 2 + u 2 ā a ln ā” ā£ a + a 2 + u 2 u ā£ + C \int\frac{\sqrt{a^2+u^2}}u\,du=\sqrt{a^2+u^2}-a\ln\left|\frac{a+\sqrt{a^2+u^2}}u\right|+C ā« u a 2 + u 2 ā ā d u = a 2 + u 2 ā ā a ln ā u a + a 2 + u 2 ā ā ā + C ā« a 2 + u 2 u 2 ā d u = ā a 2 + u 2 u + ln ā” ( u + a 2 + u 2 ) + C \int\frac{\sqrt{a^2+u^2}}{u^2}\,du=-\frac{\sqrt{a^2+u^2}}u+\ln\left(u+\sqrt{a^2+u^2}\right)+C ā« u 2 a 2 + u 2 ā ā d u = ā u a 2 + u 2 ā ā + ln ( u + a 2 + u 2 ā ) + C ā« d u a 2 + u 2 = ln ā” ( u + a 2 + u 2 ) + C \int\frac{du}{\sqrt{a^2+u^2}}=\ln\left(u+\sqrt{a^2+u^2}\right)+C ā« a 2 + u 2 ā d u ā = ln ( u + a 2 + u 2 ā ) + C ā« u 2 ā d u a 2 + u 2 = u 2 a 2 + u 2 ā a 2 2 ln ā” ( u + a 2 + u 2 ) + C \int\frac{u^2\,du}{\sqrt{a^2+u^2}}=\frac u2\sqrt{a^2+u^2}-\frac{a^2}2\ln\left(u+\sqrt{a^2+u^2}\right)+C ā« a 2 + u 2 ā u 2 d u ā = 2 u ā a 2 + u 2 ā ā 2 a 2 ā ln ( u + a 2 + u 2 ā ) + C ā« d u u a 2 + u 2 = ā 1 a ln ā” ā£ a 2 + u 2 + a u ā£ + C \int\frac{du}{u\sqrt{a^2+u^2}}=-\frac1a\ln\left|\frac{\sqrt{a^2+u^2}+a}{u}\right|+C ā« u a 2 + u 2 ā d u ā = ā a 1 ā ln ā u a 2 + u 2 ā + a ā ā + C ā« d u u 2 a 2 + u 2 = ā a 2 + u 2 a 2 u + C \int\frac{du}{u^2\sqrt{a^2+u^2}}=-\frac{\sqrt{a^2+u^2}}{a^2u}+C ā« u 2 a 2 + u 2 ā d u ā = ā a 2 u a 2 + u 2 ā ā + C ā« d u ( a 2 + u 2 ) 3 / 2 = u a 2 a 2 + u 2 + C \int\frac{du}{(a^2+u^2)^{3/2}}=\frac{u}{a^2\sqrt{a^2+u^2}}+C ā« ( a 2 + u 2 ) 3/2 d u ā = a 2 a 2 + u 2 ā u ā + C ā« a 2 ā u 2 ā d u = u 2 a 2 ā u 2 + a 2 2 sin ā” ā 1 u a + C \int\sqrt{a^2-u^2}\,du=\frac u2\sqrt{a^2-u^2}+\frac{a^2}2\sin^{-1}\frac ua+C ā« a 2 ā u 2 ā d u = 2 u ā a 2 ā u 2 ā + 2 a 2 ā sin ā 1 a u ā + C ā« u 2 a 2 ā u 2 ā d u = u 8 ( 2 u 2 ā a 2 ) a 2 ā u 2 + a 4 8 sin ā” ā 1 u a + C \int u^2\sqrt{a^2-u^2}\,du=\frac u8(2u^2-a^2)\sqrt{a^2-u^2}+\frac{a^4}8\sin^{-1}\frac ua+C ā« u 2 a 2 ā u 2 ā d u = 8 u ā ( 2 u 2 ā a 2 ) a 2 ā u 2 ā + 8 a 4 ā sin ā 1 a u ā + C ā« a 2 ā u 2 u ā d u = a 2 ā u 2 ā a ln ā” ā£ a + a 2 ā u 2 u ā£ + C \int\frac{\sqrt{a^2-u^2}}u\,du=\sqrt{a^2-u^2}-a\ln\left|\frac{a+\sqrt{a^2-u^2}}u\right|+C ā« u a 2 ā u 2 ā ā d u = a 2 ā u 2 ā ā a ln ā u a + a 2 ā u 2 ā ā ā + C ā« a 2 ā u 2 u 2 ā d u = ā a 2 ā u 2 u ā sin ā” ā 1 u a + C \int\frac{\sqrt{a^2-u^2}}{u^2}\,du=-\frac{\sqrt{a^2-u^2}}u-\sin^{-1}\frac ua+C ā« u 2 a 2 ā u 2 ā ā d u = ā u a 2 ā u 2 ā ā ā sin ā 1 a u ā + C ā« d u u a 2 ā u 2 = ā 1 a ln ā” ā£ a + a 2 ā u 2 u ā£ + C \int\frac{du}{u\sqrt{a^2-u^2}}=-\frac1a\ln\left|\frac{a+\sqrt{a^2-u^2}}u\right|+C ā« u a 2 ā u 2 ā d u ā = ā a 1 ā ln ā u a + a 2 ā u 2 ā ā ā + C ā« d u u 2 a 2 ā u 2 = ā 1 a 2 u a 2 ā u 2 + C \int\frac{du}{u^2\sqrt{a^2-u^2}}=-\frac1{a^2u}\sqrt{a^2-u^2}+C ā« u 2 a 2 ā u 2 ā d u ā = ā a 2 u 1 ā a 2 ā u 2 ā + C ā« ( a 2 ā u 2 ) 3 / 2 ā d u = ā u 8 ( 2 u 2 ā 5 a 2 ) a 2 ā u 2 + 3 a 4 8 sin ā” ā 1 u a + C \int(a^2-u^2)^{3/2}\,du=-\frac u8(2u^2-5a^2)\sqrt{a^2-u^2}+\frac{3a^4}8\sin^{-1}\frac ua+C ā« ( a 2 ā u 2 ) 3/2 d u = ā 8 u ā ( 2 u 2 ā 5 a 2 ) a 2 ā u 2 ā + 8 3 a 4 ā sin ā 1 a u ā + C ā« d u ( a 2 ā u 2 ) 3 / 2 = u a 2 a 2 ā u 2 + C \int\frac{du}{(a^2-u^2)^{3/2}}=\frac{u}{a^2\sqrt{a^2-u^2}}+C ā« ( a 2 ā u 2 ) 3/2 d u ā = a 2 a 2 ā u 2 ā u ā + C ā« u 2 ā a 2 ā d u = u 2 u 2 ā a 2 ā a 2 2 ln ā” ā£ u + u 2 ā a 2 ā£ + C \int\sqrt{u^2-a^2}\,du=\frac u2\sqrt{u^2-a^2}-\frac{a^2}2\ln\left|u+\sqrt{u^2-a^2}\right|+C ā« u 2 ā a 2 ā d u = 2 u ā u 2 ā a 2 ā ā 2 a 2 ā ln ā u + u 2 ā a 2 ā ā + C ā« u 2 u 2 ā a 2 ā d u = u 8 ( 2 u 2 ā a 2 ) u 2 ā a 2 ā a 4 8 ln ā” ā£ u + u 2 ā a 2 ā£ + C \int u^2\sqrt{u^2-a^2}\,du=\frac u8(2u^2-a^2)\sqrt{u^2-a^2}-\frac{a^4}8\ln\left|u+\sqrt{u^2-a^2}\right|+C ā« u 2 u 2 ā a 2 ā d u = 8 u ā ( 2 u 2 ā a 2 ) u 2 ā a 2 ā ā 8 a 4 ā ln ā u + u 2 ā a 2 ā ā + C ā« u 2 ā a 2 u ā d u = u 2 ā a 2 ā a cos ā” ā 1 ā£ a u ā£ + C \int\frac{\sqrt{u^2-a^2}}u\,du=\sqrt{u^2-a^2}-a\cos^{-1}\left|\frac au\right|+C ā« u u 2 ā a 2 ā ā d u = u 2 ā a 2 ā ā a cos ā 1 ā u a ā ā + C ā« u 2 ā a 2 u 2 ā d u = ā u 2 ā a 2 u + ln ā” ā£ u + u 2 ā a 2 ā£ + C \int\frac{\sqrt{u^2-a^2}}{u^2}\,du=-\frac{\sqrt{u^2-a^2}}u+\ln\left|u+\sqrt{u^2-a^2}\right|+C ā« u 2 u 2 ā a 2 ā ā d u = ā u u 2 ā a 2 ā ā + ln ā u + u 2 ā a 2 ā ā + C ā« d u u 2 ā a 2 = ln ā” ā£ u + u 2 ā a 2 ā£ + C \int\frac{du}{\sqrt{u^2-a^2}}=\ln\left|u+\sqrt{u^2-a^2}\right|+C ā« u 2 ā a 2 ā d u ā = ln ā u + u 2 ā a 2 ā ā + C ā« u 2 ā d u u 2 ā a 2 = u 2 u 2 ā a 2 + a 2 2 ln ā” ā£ u + u 2 ā a 2 ā£ + C \int\frac{u^2\,du}{\sqrt{u^2-a^2}}=\frac u2\sqrt{u^2-a^2}+\frac{a^2}2\ln\left|u+\sqrt{u^2-a^2}\right|+C ā« u 2 ā a 2 ā u 2 d u ā = 2 u ā u 2 ā a 2 ā + 2 a 2 ā ln ā u + u 2 ā a 2 ā ā + C ā« d u u 2 u 2 ā a 2 = u 2 ā a 2 a 2 u + C \int\frac{du}{u^2\sqrt{u^2-a^2}}=\frac{\sqrt{u^2-a^2}}{a^2u}+C ā« u 2 u 2 ā a 2 ā d u ā = a 2 u u 2 ā a 2 ā ā + C ā« d u ( u 2 ā a 2 ) 3 / 2 = ā u a 2 u 2 ā a 2 + C \int\frac{du}{(u^2-a^2)^{3/2}}=-\frac{u}{a^2\sqrt{u^2-a^2}}+C ā« ( u 2 ā a 2 ) 3/2 d u ā = ā a 2 u 2 ā a 2 ā u ā + C ā« u ā d u a + b u = 1 b 2 ( a + b u ā a ln ā” ā£ a + b u ā£ ) + C \int\frac{u\,du}{a+bu}=\frac1{b^2}\left(a+bu-a\ln|a+bu|\right)+C ā« a + b u u d u ā = b 2 1 ā ( a + b u ā a ln ā£ a + b u ā£ ) + C ā« u 2 ā d u a + b u = 1 2 b 3 [ ( a + b u ) 2 ā 4 a ( a + b u ) + 2 a 2 ln ā” ā£ a + b u ā£ ] + C \int\frac{u^2\,du}{a+bu}=\frac1{2b^3}\left[(a+bu)^2-4a(a+bu)+2a^2\ln|a+bu|\right]+C ā« a + b u u 2 d u ā = 2 b 3 1 ā [ ( a + b u ) 2 ā 4 a ( a + b u ) + 2 a 2 ln ā£ a + b u ā£ ] + C ā« d u u ( a + b u ) = 1 a ln ā” ā£ u a + b u ā£ + C \int\frac{du}{u(a+bu)}=\frac1a\ln\left|\frac{u}{a+bu}\right|+C ā« u ( a + b u ) d u ā = a 1 ā ln ā a + b u u ā ā + C ā« d u u 2 ( a + b u ) = ā 1 a u + b a 2 ln ā” ā£ a + b u u ā£ + C \int\frac{du}{u^2(a+bu)}=-\frac1{au}+\frac b{a^2}\ln\left|\frac{a+bu}{u}\right|+C ā« u 2 ( a + b u ) d u ā = ā a u 1 ā + a 2 b ā ln ā u a + b u ā ā + C ā« u ā d u ( a + b u ) 2 = a b 2 ( a + b u ) + 1 b 2 ln ā” ā£ a + b u ā£ + C \int\frac{u\,du}{(a+bu)^2}=\frac a{b^2(a+bu)}+\frac1{b^2}\ln|a+bu|+C ā« ( a + b u ) 2 u d u ā = b 2 ( a + b u ) a ā + b 2 1 ā ln ā£ a + b u ā£ + C ā« d u u ( a + b u ) 2 = 1 a ( a + b u ) ā 1 a 2 ln ā” ā£ a + b u u ā£ + C \int\frac{du}{u(a+bu)^2}=\frac1{a(a+bu)}-\frac1{a^2}\ln\left|\frac{a+bu}{u}\right|+C ā« u ( a + b u ) 2 d u ā = a ( a + b u ) 1 ā ā a 2 1 ā ln ā u a + b u ā ā + C ā« u 2 ā d u ( a + b u ) 2 = 1 b 3 ( a + b u ā a 2 a + b u ā 2 a ln ā” ā£ a + b u ā£ ) + C \int\frac{u^2\,du}{(a+bu)^2}=\frac1{b^3}\left(a+bu-\frac{a^2}{a+bu}-2a\ln|a+bu|\right)+C ā« ( a + b u ) 2 u 2 d u ā = b 3 1 ā ( a + b u ā a + b u a 2 ā ā 2 a ln ā£ a + b u ā£ ) + C ā« u a + b u ā d u = 2 15 b 2 ( 3 b u ā 2 a ) ( a + b u ) 3 / 2 + C \int u\sqrt{a+bu}\,du=\frac2{15b^2}(3bu-2a)(a+bu)^{3/2}+C ā« u a + b u ā d u = 15 b 2 2 ā ( 3 b u ā 2 a ) ( a + b u ) 3/2 + C ā« u ā d u a + b u = 2 3 b 2 ( b u ā 2 a ) a + b u + C \int\frac{u\,du}{\sqrt{a+bu}}=\frac2{3b^2}(bu-2a)\sqrt{a+bu}+C ā« a + b u ā u d u ā = 3 b 2 2 ā ( b u ā 2 a ) a + b u ā + C ā« u 2 ā d u a + b u = 2 15 b 3 ( 8 a 2 + 3 b 2 u 2 ā 4 a b u ) a + b u + C \int\frac{u^2\,du}{\sqrt{a+bu}}=\frac2{15b^3}(8a^2+3b^2u^2-4abu)\sqrt{a+bu}+C ā« a + b u ā u 2 d u ā = 15 b 3 2 ā ( 8 a 2 + 3 b 2 u 2 ā 4 ab u ) a + b u ā + C ā« d u u a + b u = 1 a ln ā” ā£ a + b u ā a a + b u + a ā£ + C ( a > 0 ) \int\frac{du}{u\sqrt{a+bu}}=\frac1{\sqrt a}\ln\left|\frac{\sqrt{a+bu}-\sqrt a}{\sqrt{a+bu}+\sqrt a}\right|+C\quad(a>0) ā« u a + b u ā d u ā = a ā 1 ā ln ā a + b u ā + a ā a + b u ā ā a ā ā ā + C ( a > 0 ) ā« a + b u u ā d u = 2 a + b u + a ā« d u u a + b u \int\frac{\sqrt{a+bu}}u\,du=2\sqrt{a+bu}+a\int\frac{du}{u\sqrt{a+bu}} ā« u a + b u ā ā d u = 2 a + b u ā + a ā« u a + b u ā d u ā ā« a + b u u 2 ā d u = ā a + b u u + b 2 ā« d u u a + b u \int\frac{\sqrt{a+bu}}{u^2}\,du=-\frac{\sqrt{a+bu}}u+\frac b2\int\frac{du}{u\sqrt{a+bu}} ā« u 2 a + b u ā ā d u = ā u a + b u ā ā + 2 b ā ā« u a + b u ā d u ā ā« u n a + b u ā d u = 2 b ( 2 n + 3 ) [ u n ( a + b u ) 3 / 2 ā n a ā« u n ā 1 a + b u ā d u ] \int u^n\sqrt{a+bu}\,du=\frac2{b(2n+3)}\left[u^n(a+bu)^{3/2}-na\int u^{n-1}\sqrt{a+bu}\,du\right] ā« u n a + b u ā d u = b ( 2 n + 3 ) 2 ā [ u n ( a + b u ) 3/2 ā na ā« u n ā 1 a + b u ā d u ] ā« u n ā d u a + b u = 2 u n a + b u b ( 2 n + 1 ) ā 2 n a b ( 2 n + 1 ) ā« u n ā 1 ā d u a + b u \int\frac{u^n\,du}{\sqrt{a+bu}}=\frac{2u^n\sqrt{a+bu}}{b(2n+1)}-\frac{2na}{b(2n+1)}\int\frac{u^{n-1}\,du}{\sqrt{a+bu}} ā« a + b u ā u n d u ā = b ( 2 n + 1 ) 2 u n a + b u ā ā ā b ( 2 n + 1 ) 2 na ā ā« a + b u ā u n ā 1 d u ā ā« d u u n a + b u = ā a + b u a ( n ā 1 ) u n ā 1 ā b ( 2 n ā 3 ) 2 a ( n ā 1 ) ā« d u u n ā 1 a + b u \int\frac{du}{u^n\sqrt{a+bu}}=-\frac{\sqrt{a+bu}}{a(n-1)u^{n-1}}-\frac{b(2n-3)}{2a(n-1)}\int\frac{du}{u^{n-1}\sqrt{a+bu}} ā« u n a + b u ā d u ā = ā a ( n ā 1 ) u n ā 1 a + b u ā ā ā 2 a ( n ā 1 ) b ( 2 n ā 3 ) ā ā« u n ā 1 a + b u ā d u ā ā« sin ā” 2 u ā d u = 1 2 u ā 1 4 sin ā” 2 u + C \int\sin^2u\,du=\frac12u-\frac14\sin2u+C ā« sin 2 u d u = 2 1 ā u ā 4 1 ā sin 2 u + C ā« cos ā” 2 u ā d u = 1 2 u + 1 4 sin ā” 2 u + C \int\cos^2u\,du=\frac12u+\frac14\sin2u+C ā« cos 2 u d u = 2 1 ā u + 4 1 ā sin 2 u + C ā« tan ā” 2 u ā d u = tan ā” u ā u + C \int\tan^2u\,du=\tan u-u+C ā« tan 2 u d u = tan u ā u + C ā« cot ā” 2 u ā d u = ā cot ā” u ā u + C \int\cot^2u\,du=-\cot u-u+C ā« cot 2 u d u = ā cot u ā u + C ā« sin ā” 3 u ā d u = ā 1 3 ( 2 + sin ā” 2 u ) cos ā” u + C \int\sin^3u\,du=-\frac13(2+\sin^2u)\cos u+C ā« sin 3 u d u = ā 3 1 ā ( 2 + sin 2 u ) cos u + C ā« cos ā” 3 u ā d u = 1 3 ( 2 + cos ā” 2 u ) sin ā” u + C \int\cos^3u\,du=\frac13(2+\cos^2u)\sin u+C ā« cos 3 u d u = 3 1 ā ( 2 + cos 2 u ) sin u + C ā« tan ā” 3 u ā d u = 1 2 tan ā” 2 u + ln ā” ā£ cos ā” u ā£ + C \int\tan^3u\,du=\frac12\tan^2u+\ln|\cos u|+C ā« tan 3 u d u = 2 1 ā tan 2 u + ln ā£ cos u ā£ + C ā« cot ā” 3 u ā d u = ā 1 2 cot ā” 2 u ā ln ā” ā£ sin ā” u ā£ + C \int\cot^3u\,du=-\frac12\cot^2u-\ln|\sin u|+C ā« cot 3 u d u = ā 2 1 ā cot 2 u ā ln ā£ sin u ā£ + C ā« sec ā” 3 u ā d u = 1 2 sec ā” u tan ā” u + 1 2 ln ā” ā£ sec ā” u + tan ā” u ā£ + C \int\sec^3u\,du=\frac12\sec u\tan u+\frac12\ln|\sec u+\tan u|+C ā« sec 3 u d u = 2 1 ā sec u tan u + 2 1 ā ln ā£ sec u + tan u ā£ + C ā« csc ā” 3 u ā d u = ā 1 2 csc ā” u cot ā” u + 1 2 ln ā” ā£ csc ā” u ā cot ā” u ā£ + C \int\csc^3u\,du=-\frac12\csc u\cot u+\frac12\ln|\csc u-\cot u|+C ā« csc 3 u d u = ā 2 1 ā csc u cot u + 2 1 ā ln ā£ csc u ā cot u ā£ + C ā« sin ā” n u ā d u = ā 1 n sin ā” n ā 1 u cos ā” u + n ā 1 n ā« sin ā” n ā 2 u ā d u \int\sin^nu\,du=-\frac1n\sin^{n-1}u\cos u+\frac{n-1}{n}\int\sin^{n-2}u\,du ā« sin n u d u = ā n 1 ā sin n ā 1 u cos u + n n ā 1 ā ā« sin n ā 2 u d u ā« cos ā” n u ā d u = 1 n cos ā” n ā 1 u sin ā” u + n ā 1 n ā« cos ā” n ā 2 u ā d u \int\cos^nu\,du=\frac1n\cos^{n-1}u\sin u+\frac{n-1}{n}\int\cos^{n-2}u\,du ā« cos n u d u = n 1 ā cos n ā 1 u sin u + n n ā 1 ā ā« cos n ā 2 u d u ā« tan ā” n u ā d u = 1 n ā 1 tan ā” n ā 1 u ā ā« tan ā” n ā 2 u ā d u \int\tan^nu\,du=\frac1{n-1}\tan^{n-1}u-\int\tan^{n-2}u\,du ā« tan n u d u = n ā 1 1 ā tan n ā 1 u ā ā« tan n ā 2 u d u ā« cot ā” n u ā d u = ā 1 n ā 1 cot ā” n ā 1 u ā ā« cot ā” n ā 2 u ā d u \int\cot^nu\,du=-\frac1{n-1}\cot^{n-1}u-\int\cot^{n-2}u\,du ā« cot n u d u = ā n ā 1 1 ā cot n ā 1 u ā ā« cot n ā 2 u d u ā« sec ā” n u ā d u = 1 n ā 1 tan ā” u sec ā” n ā 2 u + n ā 2 n ā 1 ā« sec ā” n ā 2 u ā d u \int\sec^nu\,du=\frac1{n-1}\tan u\sec^{n-2}u+\frac{n-2}{n-1}\int\sec^{n-2}u\,du ā« sec n u d u = n ā 1 1 ā tan u sec n ā 2 u + n ā 1 n ā 2 ā ā« sec n ā 2 u d u ā« csc ā” n u ā d u = ā 1 n ā 1 cot ā” u csc ā” n ā 2 u + n ā 2 n ā 1 ā« csc ā” n ā 2 u ā d u \int\csc^nu\,du=-\frac1{n-1}\cot u\csc^{n-2}u+\frac{n-2}{n-1}\int\csc^{n-2}u\,du ā« csc n u d u = ā n ā 1 1 ā cot u csc n ā 2 u + n ā 1 n ā 2 ā ā« csc n ā 2 u d u ā« sin ā” a u sin ā” b u ā d u = sin ā” ( a ā b ) u 2 ( a ā b ) ā sin ā” ( a + b ) u 2 ( a + b ) + C \int\sin au\sin bu\,du=\frac{\sin(a-b)u}{2(a-b)}-\frac{\sin(a+b)u}{2(a+b)}+C ā« sin a u sin b u d u = 2 ( a ā b ) s i n ( a ā b ) u ā ā 2 ( a + b ) s i n ( a + b ) u ā + C ā« cos ā” a u cos ā” b u ā d u = sin ā” ( a ā b ) u 2 ( a ā b ) + sin ā” ( a + b ) u 2 ( a + b ) + C \int\cos au\cos bu\,du=\frac{\sin(a-b)u}{2(a-b)}+\frac{\sin(a+b)u}{2(a+b)}+C ā« cos a u cos b u d u = 2 ( a ā b ) s i n ( a ā b ) u ā + 2 ( a + b ) s i n ( a + b ) u ā + C ā« sin ā” a u cos ā” b u ā d u = ā cos ā” ( a ā b ) u 2 ( a ā b ) ā cos ā” ( a + b ) u 2 ( a + b ) + C \int\sin au\cos bu\,du=-\frac{\cos(a-b)u}{2(a-b)}-\frac{\cos(a+b)u}{2(a+b)}+C ā« sin a u cos b u d u = ā 2 ( a ā b ) c o s ( a ā b ) u ā ā 2 ( a + b ) c o s ( a + b ) u ā + C ā« u sin ā” u ā d u = sin ā” u ā u cos ā” u + C \int u\sin u\,du=\sin u-u\cos u+C ā« u sin u d u = sin u ā u cos u + C ā« u cos ā” u ā d u = cos ā” u + u sin ā” u + C \int u\cos u\,du=\cos u+u\sin u+C ā« u cos u d u = cos u + u sin u + C ā« u n sin ā” u ā d u = ā u n cos ā” u + n ā« u n ā 1 cos ā” u ā d u \int u^n\sin u\,du=-u^n\cos u+n\int u^{n-1}\cos u\,du ā« u n sin u d u = ā u n cos u + n ā« u n ā 1 cos u d u ā« u n cos ā” u ā d u = u n sin ā” u ā n ā« u n ā 1 sin ā” u ā d u \int u^n\cos u\,du=u^n\sin u-n\int u^{n-1}\sin u\,du ā« u n cos u d u = u n sin u ā n ā« u n ā 1 sin u d u ā« sin ā” n u cos ā” m u ā d u = ā sin ā” n ā 1 u cos ā” m + 1 u n + m + n ā 1 n + m ā« sin ā” n ā 2 u cos ā” m u ā d u = sin ā” n + 1 u cos ā” m ā 1 u n + m + m ā 1 n + m ā« sin ā” n u cos ā” m ā 2 u ā d u
\begin{aligned}
\int\sin^nu\cos^mu\,du&=-\frac{\sin^{n-1}u\cos^{m+1}u}{n+m}+\frac{n-1}{n+m}\int\sin^{n-2}u\cos^mu\,du\\&=\frac{\sin^{n+1}u\cos^{m-1}u}{n+m}+\frac{m-1}{n+m}\int\sin^nu\cos^{m-2}u\,du
\end{aligned}
ā« sin n u cos m u d u ā = ā n + m sin n ā 1 u cos m + 1 u ā + n + m n ā 1 ā ā« sin n ā 2 u cos m u d u = n + m sin n + 1 u cos m ā 1 u ā + n + m m ā 1 ā ā« sin n u cos m ā 2 u d u ā ā« sin ā” ā 1 u ā d u = u sin ā” ā 1 u + 1 ā u 2 + C \int\sin^{-1}u\,du=u\sin^{-1}u+\sqrt{1-u^2}+C ā« sin ā 1 u d u = u sin ā 1 u + 1 ā u 2 ā + C ā« cos ā” ā 1 u ā d u = u cos ā” ā 1 u ā 1 ā u 2 + C \int\cos^{-1}u\,du=u\cos^{-1}u-\sqrt{1-u^2}+C ā« cos ā 1 u d u = u cos ā 1 u ā 1 ā u 2 ā + C ā« tan ā” ā 1 u ā d u = u tan ā” ā 1 u ā 1 2 ln ā” ( 1 + u 2 ) + C \int\tan^{-1}u\,du=u\tan^{-1}u-\frac12\ln(1+u^2)+C ā« tan ā 1 u d u = u tan ā 1 u ā 2 1 ā ln ( 1 + u 2 ) + C ā« u sin ā” ā 1 u ā d u = 2 u 2 ā 1 4 sin ā” ā 1 u + u 1 ā u 2 4 + C \int u\sin^{-1}u\,du=\frac{2u^2-1}{4}\sin^{-1}u+\frac{u\sqrt{1-u^2}}4+C ā« u sin ā 1 u d u = 4 2 u 2 ā 1 ā sin ā 1 u + 4 u 1 ā u 2 ā ā + C ā« u cos ā” ā 1 u ā d u = 2 u 2 ā 1 4 cos ā” ā 1 u ā u 1 ā u 2 4 + C \int u\cos^{-1}u\,du=\frac{2u^2-1}{4}\cos^{-1}u-\frac{u\sqrt{1-u^2}}4+C ā« u cos ā 1 u d u = 4 2 u 2 ā 1 ā cos ā 1 u ā 4 u 1 ā u 2 ā ā + C ā« u tan ā” ā 1 u ā d u = u 2 + 1 2 tan ā” ā 1 u ā u 2 + C \int u\tan^{-1}u\,du=\frac{u^2+1}{2}\tan^{-1}u-\frac u2+C ā« u tan ā 1 u d u = 2 u 2 + 1 ā tan ā 1 u ā 2 u ā + C ā« u n sin ā” ā 1 u ā d u = 1 n + 1 [ u n + 1 sin ā” ā 1 u ā ā« u n + 1 ā d u 1 ā u 2 ] , n ā ā 1 \int u^n\sin^{-1}u\,du=\frac1{n+1}\left[u^{n+1}\sin^{-1}u-\int\frac{u^{n+1}\,du}{\sqrt{1-u^2}}\right],\quad n\ne-1 ā« u n sin ā 1 u d u = n + 1 1 ā [ u n + 1 sin ā 1 u ā ā« 1 ā u 2 ā u n + 1 d u ā ] , n ī = ā 1 ā« u n cos ā” ā 1 u ā d u = 1 n + 1 [ u n + 1 cos ā” ā 1 u + ā« u n + 1 ā d u 1 ā u 2 ] , n ā ā 1 \int u^n\cos^{-1}u\,du=\frac1{n+1}\left[u^{n+1}\cos^{-1}u+\int\frac{u^{n+1}\,du}{\sqrt{1-u^2}}\right],\quad n\ne-1 ā« u n cos ā 1 u d u = n + 1 1 ā [ u n + 1 cos ā 1 u + ā« 1 ā u 2 ā u n + 1 d u ā ] , n ī = ā 1 ā« u n tan ā” ā 1 u ā d u = 1 n + 1 [ u n + 1 tan ā” ā 1 u ā ā« u n + 1 ā d u 1 + u 2 ] , n ā ā 1 \int u^n\tan^{-1}u\,du=\frac1{n+1}\left[u^{n+1}\tan^{-1}u-\int\frac{u^{n+1}\,du}{1+u^2}\right],\quad n\ne-1 ā« u n tan ā 1 u d u = n + 1 1 ā [ u n + 1 tan ā 1 u ā ā« 1 + u 2 u n + 1 d u ā ] , n ī = ā 1 ā« u e a u ā d u = 1 a 2 ( a u ā 1 ) e a u + C \int ue^{au}\,du=\frac1{a^2}(au-1)e^{au}+C ā« u e a u d u = a 2 1 ā ( a u ā 1 ) e a u + C ā« u n e a u ā d u = 1 a e a u u n ā n a ā« u n ā 1 e a u ā d u \int u^ne^{au}\,du=\frac1ae^{au}u^n-\frac na\int u^{n-1}e^{au}\,du ā« u n e a u d u = a 1 ā e a u u n ā a n ā ā« u n ā 1 e a u d u ā« e a u sin ā” b u ā d u = e a u a 2 + b 2 ( a sin ā” b u ā b cos ā” b u ) + C \int e^{au}\sin bu\,du=\frac{e^{au}}{a^2+b^2}(a\sin bu-b\cos bu)+C ā« e a u sin b u d u = a 2 + b 2 e a u ā ( a sin b u ā b cos b u ) + C ā« e a u cos ā” b u ā d u = e a u a 2 + b 2 ( a cos ā” b u + b sin ā” b u ) + C \int e^{au}\cos bu\,du=\frac{e^{au}}{a^2+b^2}(a\cos bu+b\sin bu)+C ā« e a u cos b u d u = a 2 + b 2 e a u ā ( a cos b u + b sin b u ) + C ā« ln ā” u ā d u = u ln ā” u ā u + C \int\ln u\,du=u\ln u-u+C ā« ln u d u = u ln u ā u + C ā« u n ln ā” u ā d u = u n + 1 ( n + 1 ) 2 [ ( n + 1 ) ln ā” u ā 1 ] + C \int u^n\ln u\,du=\frac{u^{n+1}}{(n+1)^2}\big[(n+1)\ln u-1\big]+C ā« u n ln u d u = ( n + 1 ) 2 u n + 1 ā [ ( n + 1 ) ln u ā 1 ] + C ā« 1 u ln ā” u ā d u = ln ā” ā£ ln ā” u ā£ + C \int\frac1{u\ln u}\,du=\ln|\ln u|+C ā« u l n u 1 ā d u = ln ā£ ln u ā£ + C ā« sinh ā” u ā d u = cosh ā” u + C \int\sinh u\,du=\cosh u+C ā« sinh u d u = cosh u + C ā« cosh ā” u ā d u = sinh ā” u + C \int\cosh u\,du=\sinh u+C ā« cosh u d u = sinh u + C ā« tanh ā” u ā d u = ln ā” cosh ā” u + C \int\tanh u\,du=\ln\cosh u+C ā« tanh u d u = ln cosh u + C ā« coth ā” u ā d u = ln ā” ā£ sinh ā” u ā£ + C \int\coth u\,du=\ln|\sinh u|+C ā« coth u d u = ln ā£ sinh u ā£ + C ā« sech ā” u ā d u = tan ā” ā 1 ā£ sinh ā” u ā£ + C \int\operatorname{sech}u\,du=\tan^{-1}|\sinh u|+C ā« sech u d u = tan ā 1 ā£ sinh u ā£ + C ā« csch ā” u ā d u = ln ā” ā£ tanh ā” 1 2 u ā£ + C \int\operatorname{csch}u\,du=\ln\left|\tanh\frac12u\right|+C ā« csch u d u = ln ā tanh 2 1 ā u ā + C ā« sech ā” 2 u ā d u = tanh ā” u + C \int\operatorname{sech}^2u\,du=\tanh u+C ā« sech 2 u d u = tanh u + C ā« csch ā” 2 u ā d u = ā coth ā” u + C \int\operatorname{csch}^2u\,du=-\coth u+C ā« csch 2 u d u = ā coth u + C ā« sech ā” u tanh ā” u ā d u = ā sech ā” u + C \int\operatorname{sech}u\tanh u\,du=-\operatorname{sech}u+C ā« sech u tanh u d u = ā sech u + C ā« csch ā” u coth ā” u ā d u = ā csch ā” u + C \int\operatorname{csch}u\coth u\,du=-\operatorname{csch}u+C ā« csch u coth u d u = ā csch u + C ā« 2 a u ā u 2 ā d u = u ā a 2 2 a u ā u 2 + a 2 2 cos ā” ā 1 ( a ā u a ) + C \int\sqrt{2au-u^2}\,du=\frac{u-a}{2}\sqrt{2au-u^2}+\frac{a^2}{2}\cos^{-1}\left(\frac{a-u}{a}\right)+C ā« 2 a u ā u 2 ā d u = 2 u ā a ā 2 a u ā u 2 ā + 2 a 2 ā cos ā 1 ( a a ā u ā ) + C ā« u 2 a u ā u 2 ā d u = 2 u 2 ā a u ā 3 a 2 6 2 a u ā u 2 + a 3 2 cos ā” ā 1 ( a ā u a ) + C \int u\sqrt{2au-u^2}\,du=\frac{2u^2-au-3a^2}{6}\sqrt{2au-u^2}+\frac{a^3}{2}\cos^{-1}\left(\frac{a-u}{a}\right)+C ā« u 2 a u ā u 2 ā d u = 6 2 u 2 ā a u ā 3 a 2 ā 2 a u ā u 2 ā + 2 a 3 ā cos ā 1 ( a a ā u ā ) + C ā« 2 a u ā u 2 u ā d u = 2 a u ā u 2 + a cos ā” ā 1 ( a ā u a ) + C \int\frac{\sqrt{2au-u^2}}u\,du=\sqrt{2au-u^2}+a\cos^{-1}\left(\frac{a-u}{a}\right)+C ā« u 2 a u ā u 2 ā ā d u = 2 a u ā u 2 ā + a cos ā 1 ( a a ā u ā ) + C ā« 2 a u ā u 2 u 2 ā d u = ā 2 2 a u ā u 2 u ā cos ā” ā 1 ( a ā u a ) + C \int\frac{\sqrt{2au-u^2}}{u^2}\,du=-\frac{2\sqrt{2au-u^2}}u-\cos^{-1}\left(\frac{a-u}{a}\right)+C ā« u 2 2 a u ā u 2 ā ā d u = ā u 2 2 a u ā u 2 ā ā ā cos ā 1 ( a a ā u ā ) + C ā« d u 2 a u ā u 2 = cos ā” ā 1 ( a ā u a ) + C \int\frac{du}{\sqrt{2au-u^2}}=\cos^{-1}\left(\frac{a-u}{a}\right)+C ā« 2 a u ā u 2 ā d u ā = cos ā 1 ( a a ā u ā ) + C ā« u ā d u 2 a u ā u 2 = ā 2 a u ā u 2 + a cos ā” ā 1 ( a ā u a ) + C \int\frac{u\,du}{\sqrt{2au-u^2}}=-\sqrt{2au-u^2}+a\cos^{-1}\left(\frac{a-u}{a}\right)+C ā« 2 a u ā u 2 ā u d u ā = ā 2 a u ā u 2 ā + a cos ā 1 ( a a ā u ā ) + C ā« u 2 ā d u 2 a u ā u 2 = ā u + 3 a 2 2 a u ā u 2 + 3 a 2 2 cos ā” ā 1 ( a ā u a ) + C \int\frac{u^2\,du}{\sqrt{2au-u^2}}=-\frac{u+3a}{2}\sqrt{2au-u^2}+\frac{3a^2}{2}\cos^{-1}\left(\frac{a-u}{a}\right)+C ā« 2 a u ā u 2 ā u 2 d u ā = ā 2 u + 3 a ā 2 a u ā u 2 ā + 2 3 a 2 ā cos ā 1 ( a a ā u ā ) + C ā« d u u 2 a u ā u 2 = ā 2 a u ā u 2 a u + C \int\frac{du}{u\sqrt{2au-u^2}}=-\frac{\sqrt{2au-u^2}}{au}+C ā« u 2 a u ā u 2 ā d u ā = ā a u 2 a u ā u 2 ā ā + C How to Use This Table This reference page contains common antiderivatives and integration formulas used throughout Calculus. Most formulas are written using the variable u u u
Many integrals can be solved directly by matching them to one of the forms in this table. Others may require algebraic simplification, substitution, or integration techniques such as integration by parts or trigonometric identities before applying a formula.
Integration by Parts is shown in formula 1 and is commonly used for products such as x e x x e^x x e x x sin ā” x x\sin x x sin x Reduction Formulas appear throughout the trigonometric and recursive sections (such as formulas 73-78 and 84-86). These rewrite a difficult integral in terms of a simpler one.Integrals involving square roots like a 2 ā u 2 \sqrt{a^2-u^2} a 2 ā u 2 ā u 2 ā a 2 \sqrt{u^2-a^2} u 2 ā a 2 ā Logarithmic results commonly appear when integrating rational expressions, especially forms involving 1 u \frac1u u 1 ā To see a comprehensive list of derivatives, check out our Table of Derivatives
Additional Notes The constant C C C Unless otherwise stated, constants such as a a a b b b n n n Expressions like sin ā” ā 1 u \sin^{-1}u sin ā 1 u Absolute values appear inside logarithms because antiderivatives must remain valid for both positive and negative inputs where defined. Some formulas include conditions such as a > 0 a>0 a > 0 n ā ā 1 n\ne-1 n ī = ā 1 