Welcome

Click on any integral to navigate to its interactive counterpart.

∫abxndx=1n+1xn+1\int_a^b x^n dx = \frac{1}{n+1}x^{n+1}
∫ab1xdx=ln⁡(x)\int_a^b \frac{1}{x} dx = \ln(x)
∫ab1Ax+Bdx=1Aln⁡∣Ax+B∣\int_a^b \frac{1}{Ax+B} dx = \frac{1}{A}\ln\bigr| Ax+B \bigr|
∫ab(x+c)ndx=(x+c)n+1n+1\int_a^b (x+c)^n dx = \frac{(x+c)^{n+1}}{n+1}
∫ab1x+cdx=ln⁡(x+c)\int_a^b \frac{1}{x+c} dx = \ln(x+c)
∫ab(x+c)ndx=(x+c)n+1n+1\int_a^b (x+c)^n dx = \frac{(x+c)^{n+1}}{n+1}
∫abx(x+c)ndx=(x+c)n+1((n+1)x−c)(n+1)(n+2)\int_a^b x(x+c)^n dx = \frac{(x+c)^{n+1}((n+1)x-c)}{(n+1)(n+2)}
∫ab11+x2dx=tan⁡−1(x)\int_a^b \frac{1}{1+x^2} dx = \tan^{-1}(x)
∫ab1c2+x2dx=1ctan⁡−1(xc)\int_a^b \frac{1}{c^2+x^2} dx = \frac{1}{c}\tan^{-1}\biggr(\frac{x}{c}\biggr)
∫abxc2+x2dx=12ln⁡∣c2+x2∣\int_a^b \frac{x}{c^2+x^2} dx = \frac{1}{2}\ln{\bigr| c^2 + x^2 \bigr|}
∫abx2c2+x2dx=x−ctan⁡−1(xc)\int_a^b \frac{x^2}{c^2+x^2} dx = x-c\tan^{-1}\biggr(\frac{x}{c}\biggr)
∫abx3c2+x2dx=12x2−12c2ln⁡∣c2+x2∣\int_a^b \frac{x^3}{c^2+x^2} dx = \frac{1}{2}x^2-\frac{1}{2}c^2\ln{\bigr| c^2 + x^2 \bigr|}
∫ab1Ax2+Bx+Cdx=24AC−B2tan⁡−1(2Ax+B4AC+B2)\int_a^b \frac{1}{Ax^2+Bx+C} dx = \frac{2}{\sqrt{4AC-B^2}}\tan^{-1}\biggr(\frac{2Ax+B}{\sqrt{4AC+B^2}}\biggr)
∫abxx+c dx=23(x−2c)x+c\int_a^b \frac{x}{\sqrt{x+c}}\textnormal{ }dx = \frac{2}{3}(x-2c)\sqrt{x+c}
∫abxc−x dx=−x(c−x)−ctan⁡−1(x(c−x)x−c)\int_a^b \sqrt{\frac{x}{c-x}}\textnormal{ }dx = -\sqrt{x(c-x)}-c\tan^{-1}\biggr(\frac{\sqrt{x(c-x)}}{x-c}\biggr)
∫abxc+x dx=x(c+x)−cln⁡(x+x+c)\int_a^b \sqrt{\frac{x}{c+x}}\textnormal{ }dx = \sqrt{x(c+x)}-c\ln\bigr(\sqrt{x}+\sqrt{x+c}\bigr)
∫abx2+c2 dx=12xx2+c2+12c2ln⁡∣x+x2+c2∣\int_a^b \sqrt{x^2+c^2}\textnormal{ }dx = \frac{1}{2}x\sqrt{x^2 + c^2}+ \frac{1}{2}c^2\ln{\biggr|x+\sqrt{x^2 + c^2}\biggr|}
∫abc2−x2 dx=12xc2−x2+12c2tan⁡−1(xc2−x2)\int_a^b \sqrt{c^2-x^2}\textnormal{ }dx = \frac{1}{2}x\sqrt{c^2-x^2}+ \frac{1}{2}c^2\tan^{-1}\biggr(\frac{x}{\sqrt{c^2-x^2}}\biggr)
∫ab1x2+c2 dx=ln⁡∣x+x2+c2∣\int_a^b \frac{1}{\sqrt{x^2+c^2}}\textnormal{ }dx = \ln{\biggr|x+\sqrt{x^2+c^2}\biggr|}
∫ab1c2−x2 dx=sin⁡−1(xc)\int_a^b \frac{1}{\sqrt{c^2-x^2}}\textnormal{ }dx = \sin^{-1}\biggr(\frac{x}{c}\biggr)
∫abx2x2+c2 dx=12xx2+c2−12c2ln⁡∣x+x2+c2∣\int_a^b \frac{x^2}{\sqrt{x^2+c^2}}\textnormal{ }dx = \frac{1}{2}x\sqrt{x^2+c^2}- \frac{1}{2}c^2\ln{\biggr|x+\sqrt{x^2+c^2}\biggr|}
∫ab1(c2+x2)3/2 dx=xc2c2+x2\int_a^b \frac{1}{(c^2+x^2)^{3/2}}\textnormal{ }dx = \frac{x}{c^2\sqrt{c^2+x^2}}
∫abln⁡(cx)dx=xln⁡(cx)−x\int_a^b \ln(cx) dx = x\ln(cx)-x
∫abxnln⁡(cx)dx=xn+1(ln⁡(x)n+1−1(n+1)2)\int_a^b x^n\ln(cx) dx = x^{n+1}\biggr(\frac{\ln(x)}{n+1}-\frac{1}{(n+1)^2}\biggr)
∫abln⁡(cx)xdx=12(ln⁡(cx))2\int_a^b \frac{\ln(cx)}{x} dx = \frac{1}{2}\bigr(\ln(cx)\bigr)^2
∫abln⁡(cx)x2dx=1x−ln⁡(cx)x\int_a^b \frac{\ln(cx)}{x^2} dx = \frac{1}{x}-\frac{\ln(cx)}{x}
∫abln⁡(x2+c2)dx=xln⁡(x2+c2)+2ctan⁡−1(xc)−2x\int_a^b \ln\bigr(x^2+c^2\bigr) dx = x\ln\bigr(x^2+c^2\bigr)+2c\tan^{-1}\biggr(\frac{x}{c}\biggr)-2x
∫abln⁡(x2−c2)dx=xln⁡(x2−c2)+cln⁡(x+cx−c)−2x\int_a^b \ln\bigr(x^2-c^2\bigr) dx = x\ln\bigr(x^2-c^2\bigr)+c\ln\biggr(\frac{x+c}{x-c}\biggr)-2x
∫abecxdx=1cecx\int_a^b e^{cx} dx = \frac{1}{c}e^{cx}
∫abxecxdx=(xc−1c2)ecx\int_a^b xe^{cx} dx = (\frac{x}{c}-\frac{1}{c^2})e^{cx}
∫abx2ecxdx=(x2c−2xc2+2c3)ecx\int_a^b x^{2}e^{cx} dx = (\frac{x^2}{c}-\frac{2x}{c^2}+\frac{2}{c^3})e^{cx}
∫absin⁡(cx)dx=−1ccos⁡(cx)\int_a^b \sin(cx) dx = -\frac{1}{c}\cos(cx)
∫absin⁡2(cx)dx=x2−sin⁡(2cx)4c\int_a^b \sin^{2}(cx) dx = \frac{x}{2}-\frac{\sin(2cx)}{4c}
∫absin⁡3(cx)dx=3cos⁡(cx)4c−cos⁡(3cx)12c\int_a^b \sin^{3}(cx) dx = \frac{3\cos(cx)}{4c}-\frac{\cos(3cx)}{12c}
∫abcos⁡(cx)dx=1csin⁡(cx)\int_a^b \cos(cx) dx = \frac{1}{c}\sin(cx)
∫abcos⁡2(cx)dx=x2+sin⁡(2cx)4c\int_a^b \cos^{2}(cx) dx = \frac{x}{2}+\frac{\sin(2cx)}{4c}
∫abcos⁡3(cx)dx=3sin⁡(cx)4c−sin⁡(3cx)12c\int_a^b \cos^{3}(cx) dx = \frac{3\sin(cx)}{4c}-\frac{\sin(3cx)}{12c}
∫absin⁡(cx)cos⁡(cx)dx=sin⁡2(cx)2c\int_a^b \sin(cx)\cos(cx) dx = \frac{\sin^2(cx)}{2c}
∫absin⁡(Ax)cos⁡(Bx)dx=−cos⁡[(A+B)x]2(A+B)−cos⁡[(A−B)x]2(A−B)\int_a^b \sin(Ax)\cos(Bx) dx = -\frac{\cos[(A+B)x]}{2(A+B)}-\frac{\cos[(A-B)x]}{2(A-B)}
∫absin⁡2(Ax)cos⁡(Bx)dx=−sin⁡[(2A−B)x]4(2A−B)+sin⁡(Bx)2B−sin⁡[(2A+B)x]4(2A+B)\int_a^b \sin^{2}(Ax)\cos(Bx) dx = -\frac{\sin[(2A-B)x]}{4(2A-B)}+\frac{\sin(Bx)}{2B}-\frac{\sin[(2A+B)x]}{4(2A+B)}
∫absin⁡(Ax)cos⁡2(Bx)dx=cos⁡[(2A−B)x]4(2A−B)−cos⁡(Bx)2B−cos⁡[(2A+B)x]4(2A+B)\int_a^b \sin(Ax)\cos^{2}(Bx) dx = \frac{\cos[(2A-B)x]}{4(2A-B)}-\frac{\cos(Bx)}{2B}-\frac{\cos[(2A+B)x]}{4(2A+B)}

Please report any errors or post feature requests here, thanks!