∫ 0 d x = C {\displaystyle ~\int \!0\,dx=C} (nguyên hàm của 0 là hằng số; tích phân xác định của 0 lấy trong bất kì khoảng nào thì bằng 0) ∫ a d x = a x + C {\displaystyle ~\int \!a\,dx=ax+C} ∫ x n d x = { x n + 1 n + 1 + C , n ≠ − 1 ln | x | + C , n = − 1 {\displaystyle ~\int \!x^{n}\,dx={\begin{cases}{\frac {x^{n+1}}{n+1}}+C,&n\neq -1\\\ln \left|x\right|+C,&n=-1\end{cases}}} ∫ d x a 2 + x 2 = 1 a arctg x a + C = − 1 a arcctg x a + C {\displaystyle \int \!{dx \over {a^{2}+x^{2}}}={1 \over a}\,\operatorname {arctg} \,{\frac {x}{a}}+C=-{1 \over a}\,\operatorname {arcctg} \,{\frac {x}{a}}+C}
Chứng minh
Lấy đạo hàm vế phải:
d d x ( 1 a arctg x a + C ) = d d x ( 1 a arctg x a ) = 1 a ⋅ d d ( x a ) ( arctg x a ) ⋅ d d x ( x a ) = 1 a ⋅ 1 1 + x 2 a 2 ⋅ 1 a = 1 a 2 ⋅ a 2 + x 2 a 2 = 1 a 2 + x 2 {\displaystyle {d \over dx}\,\left({1 \over a}\,\operatorname {arctg} \,{\frac {x}{a}}+C\right)={d \over dx}\,\left({1 \over a}\,\operatorname {arctg} \,{\frac {x}{a}}\right)={\frac {1}{a}}\cdot {d \over d\left({x \over a}\right)}\left(\operatorname {arctg} {\frac {x}{a}}\right)\cdot {d \over dx}\left({x \over a}\right)={\frac {1}{a}}\cdot {\frac {1}{1+{\frac {x^{2}}{a^{2}}}}}\cdot {\frac {1}{a}}={\frac {1}{a^{2}\cdot {\frac {a^{2}+x^{2}}{a^{2}}}}}={\frac {1}{a^{2}+x^{2}}}}
∫ d x x 2 − a 2 = 1 2 a ln | x − a x + a | + C {\displaystyle \int \!{dx \over {x^{2}-a^{2}}}={1 \over 2a}\ln \left|{x-a \over {x+a}}\right|+C}
∫ ln x d x = x ln x − x + C {\displaystyle \int \!\ln {x}\,dx=x\ln {x}-x+C} ∫ d x x ln x = ln | ln x | + C {\displaystyle \int {\frac {dx}{x\ln x}}=\ln |\ln x|+C} ∫ log b x d x = x log b x − x log b e + C = x ln x − 1 ln b + C {\displaystyle \int \!\log _{b}{x}\,dx=x\log _{b}{x}-x\log _{b}{e}+C=x{\frac {\ln {x}-1}{\ln b}}+C}
∫ e x d x = e x + C {\displaystyle \int \!e^{x}\,dx=e^{x}+C} ∫ a x d x = a x ln a + C {\displaystyle \int \!a^{x}\,dx={\frac {a^{x}}{\ln {a}}}+C}
∫ d x a 2 − x 2 = arcsin x a + C {\displaystyle \int \!{dx \over {\sqrt {a^{2}-x^{2}}}}=\arcsin {x \over a}+C} ∫ − d x a 2 − x 2 = arccos x a + C {\displaystyle \int \!{-dx \over {\sqrt {a^{2}-x^{2}}}}=\arccos {x \over a}+C} ∫ d x x x 2 − a 2 = 1 a arcsec | x | a + C {\displaystyle \int \!{dx \over x{\sqrt {x^{2}-a^{2}}}}={1 \over a}\,\operatorname {arcsec} \,{|x| \over a}+C} ∫ d x x 2 ± a 2 = ln | x + x 2 ± a 2 | + C {\displaystyle \int \!{dx \over {\sqrt {x^{2}\pm a^{2}}}}=\ln \left|{x+{\sqrt {x^{2}\pm a^{2}}}}\right|+C} («длинный логарифм»)
∫ sin x d x = − cos x + C {\displaystyle \int \!\sin {x}\,dx=-\cos {x}+C} ∫ cos x d x = sin x + C {\displaystyle \int \!\cos {x}\,dx=\sin {x}+C} ∫ tg x d x = − ln | cos x | + C {\displaystyle \int \!\operatorname {tg} \,{x}\,dx=-\ln {\left|\cos {x}\right|}+C} ∫ ctg x d x = ln | sin x | + C {\displaystyle \int \!\operatorname {ctg} \,{x}\,dx=\ln {\left|\sin {x}\right|}+C}
Chứng minh 2 công thức trên
∫ tg x d x = ∫ sin x cos x d x = − ∫ d ( cos x ) cos x = − ln | cos x | + C {\displaystyle \int \!\operatorname {tg} \,{x}\,dx=\int {\frac {\sin x}{\cos x}}dx=-\int {\frac {d(\cos x)}{\cos x}}=-\ln |\cos x|+C}
∫ ctg x d x = ∫ cos x sin x d x = ∫ d ( sin x ) sin x = ln | sin x | + C {\displaystyle \int \!\operatorname {ctg} \,{x}\,dx=\int {\frac {\cos x}{\sin x}}dx=\int {\frac {d(\sin x)}{\sin x}}=\ln |\sin x|+C}
∫ sec x d x = ln | sec x + tg x | + C {\displaystyle \int \!\sec {x}\,dx=\ln {\left|\sec {x}+\operatorname {tg} \,{x}\right|}+C} ∫ csc x d x = − ln | csc x + ctg x | + C {\displaystyle \int \!\csc {x}\,dx=-\ln {\left|\csc {x}+\operatorname {ctg} \,{x}\right|}+C} ∫ sec 2 x d x = ∫ d x cos 2 x = tg x + C {\displaystyle \int \!\sec ^{2}x\,dx=\int \!{dx \over \cos ^{2}x}=\operatorname {tg} \,x+C} ∫ csc 2 x d x = ∫ d x sin 2 x = − ctg x + C {\displaystyle \int \!\csc ^{2}x\,dx=\int \!{dx \over \sin ^{2}x}=-\operatorname {ctg} \,x+C} ∫ sec x tg x d x = sec x + C {\displaystyle \int \!\sec {x}\,\operatorname {tg} \,{x}\,dx=\sec {x}+C} ∫ csc x ctg x d x = − csc x + C {\displaystyle \int \!\csc {x}\,\operatorname {ctg} \,{x}\,dx=-\csc {x}+C} ∫ sin 2 x d x = 1 2 ( x − sin x cos x ) + C {\displaystyle \int \!\sin ^{2}x\,dx={\frac {1}{2}}(x-\sin x\cos x)+C} ∫ cos 2 x d x = 1 2 ( x + sin x cos x ) + C {\displaystyle \int \!\cos ^{2}x\,dx={\frac {1}{2}}(x+\sin x\cos x)+C} ∫ sin n x d x = − sin n − 1 x cos x n + n − 1 n ∫ sin n − 2 x d x , n ∈ N , n ⩾ 2 {\displaystyle \int \!\sin ^{n}x\,dx=-{\frac {\sin ^{n-1}{x}\cos {x}}{n}}+{\frac {n-1}{n}}\int \!\sin ^{n-2}{x}\,dx,n\in \mathbb {N} ,n\geqslant 2} ∫ cos n x d x = cos n − 1 x sin x n + n − 1 n ∫ cos n − 2 x d x , n ∈ N , n ⩾ 2 {\displaystyle \int \!\cos ^{n}x\,dx={\frac {\cos ^{n-1}{x}\sin {x}}{n}}+{\frac {n-1}{n}}\int \!\cos ^{n-2}{x}\,dx,n\in \mathbb {N} ,n\geqslant 2} ∫ arctg x d x = x arctg x − 1 2 ln ( 1 + x 2 ) + C {\displaystyle \int \!\operatorname {arctg} \,{x}\,dx=x\,\operatorname {arctg} \,{x}-{\frac {1}{2}}\ln {\left(1+x^{2}\right)}+C}
∫ sh x d x = ch x + C {\displaystyle \int \operatorname {sh} \,x\,dx=\operatorname {ch} \,x+C} ∫ ch x d x = sh x + C {\displaystyle \int \operatorname {ch} \,x\,dx=\operatorname {sh} \,x+C} ∫ d x ch 2 x = th x + C {\displaystyle \int {\frac {dx}{\operatorname {ch} ^{2}\,x}}=\operatorname {th} \,x+C} ∫ d x sh 2 x = − cth x + C {\displaystyle \int {\frac {dx}{\operatorname {sh} ^{2}\,x}}=-\operatorname {cth} \,x+C} ∫ th x d x = ln | ch x | + C {\displaystyle \int \operatorname {th} \,x\,dx=\ln |\operatorname {ch} \,x|+C} ∫ csch x d x = ln | th x 2 | + C {\displaystyle \int \operatorname {csch} \,x\,dx=\ln \left|\operatorname {th} \,{x \over 2}\right|+C} ∫ sech x d x = arctg ( sh x ) + C {\displaystyle \int \operatorname {sech} \,x\,dx=\operatorname {arctg} \,(\operatorname {sh} \,x)+C} ngoài ra ∫ sech x d x = 2 arctg ( e x ) + C {\displaystyle \int \operatorname {sech} \,x\,dx=2\,\operatorname {arctg} \,(e^{x})+C} và ∫ sech x d x = 2 arctg ( th x 2 ) + C {\displaystyle \int \operatorname {sech} \,x\,dx=2\,\operatorname {arctg} \,\left(\operatorname {th} \,{\frac {x}{2}}\right)+C} ∫ cth x d x = ln | sh x | + C {\displaystyle \int \operatorname {cth} \,x\,dx=\ln |\operatorname {sh} \,x|+C}
Chứng minh
Chứng minh công thức ∫ sech x d x = arctan ( sh x ) + C {\displaystyle \int \operatorname {sech} \,x\,dx=\arctan(\operatorname {sh} \,x)+C} bằng cách lấy đạo hàm vế phải:
( arctg ( sh x ) + C ) ′ = ch x sh 2 x + 1 = ch x ch 2 x − 1 + 1 = 1 ch x = sech x {\displaystyle \left(\operatorname {arctg} (\operatorname {sh} \,x\right)+C)'={\frac {\operatorname {ch} \,x}{\operatorname {sh} ^{2}\,x+1}}={\frac {\operatorname {ch} \,x}{\operatorname {ch} ^{2}\,x-1+1}}={\frac {1}{\operatorname {ch} \,x}}=\operatorname {sech} \,x} .
Chứng minh công thức ∫ sech x d x = 2 arctg ( e x ) + C {\displaystyle \int \operatorname {sech} \,x\,dx=2\operatorname {arctg} (e^{x})+C} bằng cách lấy đạo hàm vế phải:
( 2 arctg ( e x ) + C ) ′ = 2 e x e 2 x + 1 = 2 e x + e − x = 1 ch x = sech x {\displaystyle \left(2\operatorname {arctg} (e^{x})+C\right)'={\frac {2e^{x}}{e^{2x}+1}}={\frac {2}{e^{x}+e^{-x}}}={\frac {1}{\operatorname {ch} \,x}}=\operatorname {sech} \,x} .
Chứng minh công thức ∫ sech x d x = 2 arctg ( th x 2 ) + C {\displaystyle \int \operatorname {sech} \,x\,dx=2\,\operatorname {arctg} \,\left(\operatorname {th} \,{\frac {x}{2}}\right)+C} bằng cách lấy đạo hàm vế phải:
( 2 arctg ( th x 2 ) + C ) ′ = sech 2 x 2 th 2 x 2 + 1 = 1 sh 2 x 2 + ch 2 x 2 = 1 ch x = sech x {\displaystyle \left(2\,\operatorname {arctg} \,\left(\operatorname {th} \,{\frac {x}{2}}\right)+C\right)'={\frac {\operatorname {sech} ^{2}{\frac {x}{2}}}{\operatorname {th} ^{2}{\frac {x}{2}}+1}}={\frac {1}{\operatorname {sh} ^{2}{\frac {x}{2}}+\operatorname {ch} ^{2}{\frac {x}{2}}}}={\frac {1}{\operatorname {ch} \,x}}=\operatorname {sech} \,x} .
