∫xαdx=xα+1α+1+C (α≠−1) \int x^\alpha dx = \dfrac{x^{\alpha+1}}{\alpha+1} + C \space (\alpha \neq -1) ∫xαdx=α+1xα+1+C (α=−1)
∫dxx=ln∣x∣+C (α=−1) \int \dfrac{dx}{x} = \ln |x| + C \space (\alpha = -1) ∫xdx=ln∣x∣+C (α=−1)
∫axdx=axlna+C (a>0,a≠1) \int a^x dx = \dfrac{a^x}{\ln a} + C \space (a > 0, a\neq 1) ∫axdx=lnaax+C (a>0,a=1)
∫exdx=ex+C \int e^x dx = e^x + C ∫exdx=ex+C
∫sinxdx=−cosx+C \int \sin x dx = -\cos x + C ∫sinxdx=−cosx+C
∫cosxdx=sinx+C \int \cos x dx = \sin x + C ∫cosxdx=sinx+C
∫1cos2xdx=tanx+C \int \dfrac{1}{\cos^2 x} dx = \tan x + C ∫cos2x1dx=tanx+C
∫1sin2x=−cotx+C \int \dfrac{1}{\sin^2 x} = -\cot x + C ∫sin2x1=−cotx+C
∫sinxcos2xdx=1cosx+C \int \dfrac{\sin x}{\cos^2 x} dx = \dfrac{1}{\cos x} + C ∫cos2xsinxdx=cosx1+C
∫cosxsin2xdx=−1sinx+C \int \dfrac{\cos x}{\sin^2 x} dx = -\dfrac{1}{\sin x} + C ∫sin2xcosxdx=−sinx1+C
∫sin2xdx=x−sinxcosx2 \int \sin^2 x dx = \dfrac{x - \sin x \cos x}{2} ∫sin2xdx=2x−sinxcosx
∫cos2xdx=x+sinxcosx2+C \int \cos^2 x dx = \dfrac{x + \sin x \cos x}{2} + C ∫cos2xdx=2x+sinxcosx+C
∫1sinxdx=ln∣1sinx−1tanx∣+C \int \dfrac{1}{\sin x} dx = \ln |\dfrac{1}{\sin x} - \dfrac{1}{\tan x}| + C ∫sinx1dx=ln∣sinx1−tanx1∣+C
∫1cosxdx=ln∣1cosx−tanx∣+C \int \dfrac{1}{\cos x} dx = \ln |\dfrac{1}{\cos x} - \tan x| + C ∫cosx1dx=ln∣cosx1−tanx∣+C
∫dxa2−x2=arcsinxa+C \int \dfrac{dx}{\sqrt{a^2 - x^2}} = \arcsin \dfrac{x}{a} + C ∫a2−x2dx=arcsinax+C
∫dxx2+a2=ln∣x+x2+a2∣+C \int \dfrac{dx}{\sqrt{x^2 + a^2}} = \ln |x + \sqrt{x^2 + a^2}| + C ∫x2+a2dx=ln∣x+x2+a2∣+C
∫dxx2−a2=ln∣x+x2−a2∣+C \int \dfrac{dx}{\sqrt{x^2 - a^2}} = \ln |x + \sqrt{x^2 - a^2}| + C ∫x2−a2dx=ln∣x+x2−a2∣+C
∫dxx2−a2=12aln∣x−ax+a∣+C \int \dfrac{dx}{x^2 - a^2} = \dfrac{1}{2a} \ln |\dfrac{x-a}{x+a}| + C ∫x2−a2dx=2a1ln∣x+ax−a∣+C
∫dxx2+a2=1aarcsin1a+C \int \dfrac{dx}{x^2 +a ^2} = \dfrac{1}{a} \arcsin \dfrac{1}{a} + C ∫x2+a2dx=a1arcsina1+C
