Inverse derivatives ∫ d x = x + C \int \mathrm{d}x = x + C ∫ d x = x + C ∫ 0 d x = C \int 0\mathrm{d}x = C ∫ 0 d x = C ∫ x m d x = x m + 1 m + 1 + C , m ≠ − 1 \int x^m\mathrm{d}x = \frac{x^{m+1}}{m+1} + C, m \neq -1 ∫ x m d x = m + 1 x m + 1 + C , m = − 1 ∫ d x x = ln ∣ x ∣ + C \int \frac{\mathrm{d}x}{x} = \ln|x| + C ∫ x d x = ln ∣ x ∣ + C ∫ cos ( x ) d x = sin ( x ) + C \int \cos(x) \mathrm{d}x = \sin(x) + C ∫ cos ( x ) d x = sin ( x ) + C ∫ sin ( x ) d x = − cos ( x ) + C \int \sin(x) \mathrm{d}x = -\cos(x) + C ∫ sin ( x ) d x = − cos ( x ) + C ∫ d x 1 + x 2 = arctg ( x ) + C = − arcctg ( x ) + C \int \frac{\mathrm{d}x}{1+x^2} = \arctg(x) + C = -\arcctg(x) + C ∫ 1 + x 2 d x = arctg ( x ) + C = − arcctg ( x ) + C ∫ d x 1 − x 2 = arcsin ( x ) + C = − arccos ( x ) + C \int \frac{\mathrm{d}x}{\sqrt{1-x^2}} = \arcsin(x) + C = -\arccos(x) + C ∫ 1 − x 2 d x = arcsin ( x ) + C = − arccos ( x ) + C ∫ a x d x = a x ln a + C \int a^x \mathrm{d}x = \frac{a^x}{\ln a} + C ∫ a x d x = ln a a x + C ∫ e x d x = e x + C \int e^x \mathrm{d}x = e^x + C ∫ e x d x = e x + C ∫ sec ( x ) 2 d x = ∫ d x cos ( x ) 2 = tg ( x ) + C \int \sec(x)^2 \mathrm{d}x = \int \frac{dx}{\cos(x)^2} = \tg(x) + C ∫ sec ( x ) 2 d x = ∫ cos ( x ) 2 d x = tg ( x ) + C ∫ cosec ( x ) 2 d x = ∫ d x sin ( x ) 2 = − ctg ( x ) + C \int \cosec(x)^2 \mathrm{d}x = \int \frac{dx}{\sin(x)^2} = -\ctg(x) + C ∫ cosec ( x ) 2 d x = ∫ sin ( x ) 2 d x = − ctg ( x ) + C ∫ sh ( x ) d x = ch ( x ) + C \int \sh(x)\mathrm{d}x = \ch(x) + C ∫ sh ( x ) d x = ch ( x ) + C ∫ ch ( x ) d x = sh ( x ) + C \int \ch(x)\mathrm{d}x = \sh(x) + C ∫ ch ( x ) d x = sh ( x ) + C ∫ d x ch ( x ) 2 = th ( x ) + C \int \frac{\mathrm{d}x}{\ch(x)^2} = \th(x) + C ∫ ch ( x ) 2 d x = th ( x ) + C ∫ d x sh ( x ) 2 = − cth ( x ) + C \int \frac{\mathrm{d}x}{\sh(x)^2} = -\cth(x) + C ∫ sh ( x ) 2 d x = − cth ( x ) + C Standart integrals ∫ d x x 2 + a 2 = 1 a arctg ( x a ) + C \int \frac{\mathrm{d}x}{x^2 + a^2} = \frac{1}{a} \arctg\Big(\frac{x}{a}\Big) + C ∫ x 2 + a 2 d x = a 1 arctg ( a x ) + C ∫ d x x 2 − a 2 = 1 2 a ln ∣ x − a x + a ∣ + C \int \frac{\mathrm{d}x}{x^2 - a^2} = \frac{1}{2a}\ln\Big|\frac{x-a}{x+a}\Big| + C ∫ x 2 − a 2 d x = 2 a 1 ln ∣ ∣ ∣ ∣ x + a x − a ∣ ∣ ∣ ∣ + C ∫ d x x 2 ± a 2 = ln ∣ x + x 2 ± a 2 ∣ + C \int \frac{\mathrm{d}x}{\sqrt{x^2 \pm a^2}} = \ln|x + \sqrt{x^2 \pm a^2}| + C ∫ x 2 ± a 2 d x = ln ∣ x + x 2 ± a 2 ∣ + C ∫ d x a 2 − x 2 = arcsin ( x a ) + C \int \frac{\mathrm{d}x}{\sqrt{a^2 - x^2}} = \arcsin \Big(\frac{x}{a}\Big) + C ∫ a 2 − x 2 d x = arcsin ( a x ) + C Copyright © 2019 — 2021 Alexander Mayorov. All rights reserved. Cookies Policy
