Integral table

Inverse derivatives

∫dx=x+C\int \mathrm{d}x = x + C ∫0dx=C\int 0\mathrm{d}x = C ∫xmdx=xm+1m+1+C,m≠−1\int x^m\mathrm{d}x = \frac{x^{m+1}}{m+1} + C, m \neq -1 ∫dxx=ln⁡∣x∣+C\int \frac{\mathrm{d}x}{x} = \ln|x| + C ∫cos⁡(x)dx=sin⁡(x)+C\int \cos(x) \mathrm{d}x = \sin(x) + C ∫sin⁡(x)dx=−cos⁡(x)+C\int \sin(x) \mathrm{d}x = -\cos(x) + C ∫dx1+x2=arctg⁡(x)+C=−arcctg⁡(x)+C\int \frac{\mathrm{d}x}{1+x^2} = \arctg(x) + C = -\arcctg(x) + C ∫dx1−x2=arcsin⁡(x)+C=−arccos⁡(x)+C\int \frac{\mathrm{d}x}{\sqrt{1-x^2}} = \arcsin(x) + C = -\arccos(x) + C ∫axdx=axln⁡a+C\int a^x \mathrm{d}x = \frac{a^x}{\ln a} + C ∫exdx=ex+C\int e^x \mathrm{d}x = e^x + C ∫sec⁡(x)2dx=∫dxcos⁡(x)2=tg⁡(x)+C\int \sec(x)^2 \mathrm{d}x = \int \frac{dx}{\cos(x)^2} = \tg(x) + C ∫cosec⁡(x)2dx=∫dxsin⁡(x)2=−ctg⁡(x)+C\int \cosec(x)^2 \mathrm{d}x = \int \frac{dx}{\sin(x)^2} = -\ctg(x) + C ∫sh⁡(x)dx=ch⁡(x)+C\int \sh(x)\mathrm{d}x = \ch(x) + C ∫ch⁡(x)dx=sh⁡(x)+C\int \ch(x)\mathrm{d}x = \sh(x) + C ∫dxch⁡(x)2=th⁡(x)+C\int \frac{\mathrm{d}x}{\ch(x)^2} = \th(x) + C ∫dxsh⁡(x)2=−cth⁡(x)+C\int \frac{\mathrm{d}x}{\sh(x)^2} = -\cth(x) + C

Standart integrals

∫dxx2+a2=1aarctg⁡(xa)+C\int \frac{\mathrm{d}x}{x^2 + a^2} = \frac{1}{a} \arctg\Big(\frac{x}{a}\Big) + C ∫dxx2−a2=12aln⁡∣x−ax+a∣+C\int \frac{\mathrm{d}x}{x^2 - a^2} = \frac{1}{2a}\ln\Big|\frac{x-a}{x+a}\Big| + C ∫dxx2±a2=ln⁡∣x+x2±a2∣+C\int \frac{\mathrm{d}x}{\sqrt{x^2 \pm a^2}} = \ln|x + \sqrt{x^2 \pm a^2}| + C ∫dxa2−x2=arcsin⁡(xa)+C\int \frac{\mathrm{d}x}{\sqrt{a^2 - x^2}} = \arcsin \Big(\frac{x}{a}\Big) + C

Copyright © 2019 — 2025 Alexander Mayorov. All rights reserved.