In base al Primo teorema fondamentale del calcolo integrale, il calcolo di suddetti integrali tramite identificazione della primitiva viene effettuato attraverso algoritmi atti a far sì che la derivata del risultato coincida con la funzione integranda. Questa pagina contiene una tavola degli integrali più comuni . Queste formule sono equivalenti a quelle presentate nella tavola delle derivate . Per altri integrali vedi Integrale § Tavole di integrali .
Qui C {\displaystyle C} costante arbitraria di integrazione che ha senso specificare solo in relazione a una specificazione del valore dell'integrale in qualche punto.
Costante:
∫ a f ( x ) d x = a ∫ f ( x ) d x {\displaystyle \int af(x)\,\mathrm {d} x=a\int f(x)\,\mathrm {d} x} Somma:
∫ [ f ( x ) + g ( x ) ] d x = ∫ f ( x ) d x + ∫ g ( x ) d x {\displaystyle \int [f(x)+g(x)]\,\mathrm {d} x=\int f(x)\,\mathrm {d} x+\int g(x)\,\mathrm {d} x} Integrazione per parti:
∫ f ( x ) g ′ ( x ) d x = f ( x ) g ( x ) − ∫ f ′ ( x ) g ( x ) d x {\displaystyle \int f(x)g'(x)\,\mathrm {d} x=f(x)g(x)-\int f'(x)g(x)\,\mathrm {d} x} ∫ d x = x + C {\displaystyle \int \,\mathrm {d} x=x+C} ∫ x a d x = x a + 1 a + 1 + C ⟺ a ≠ − 1 {\displaystyle \int x^{a}\,\mathrm {d} x={\frac {x^{a+1}}{a+1}}+C\iff a\neq -1} ∫ 1 x d x = ln | x | + C {\displaystyle \int {\frac {1}{x}}\,\mathrm {d} x=\ln {\left|x\right|}+C} ∫ x d x = 2 3 x 3 + C {\displaystyle \int {\sqrt {x}}\,\mathrm {d} x={\frac {2}{3}}{\sqrt {x^{3}}}+C} ∫ f ′ ( x ) f ( x ) d x = ln | f ( x ) | + C {\displaystyle \int {\frac {f'(x)}{f(x)}}\,\mathrm {d} x=\ln {\left|f(x)\right|}+C} ∫ f ′ ( x ) 1 + f 2 ( x ) d x = arctan f ( x ) + C {\displaystyle \int {\frac {f'(x)}{1+f^{2}(x)}}\,\mathrm {d} x=\arctan {f(x)}+C} ∫ 1 1 + x 2 d x = arctan x + C {\displaystyle \int {\frac {1}{1+x^{2}}}\,\mathrm {d} x=\arctan {x}+C} ∫ 1 a 2 + x 2 d x = 1 a arctan x a + C {\displaystyle \int {\frac {1}{a^{2}+x^{2}}}\,\mathrm {d} x={\frac {1}{a}}\arctan {\frac {x}{a}}+C} ∫ 1 a + b x 2 d x = arctan b x a a b + C {\displaystyle \int {\frac {1}{a+bx^{2}}}\,\mathrm {d} x={\frac {\arctan {\frac {{\sqrt {b}}x}{\sqrt {a}}}}{\sqrt {ab}}}+C} ∫ 1 a x 2 + b x + c d x = 1 b 2 − 4 a c ln | 2 a x + b − b 2 − 4 a c 2 a x + b + b 2 − 4 a c | + C ⟺ b 2 − 4 a c > 0 {\displaystyle \int {\frac {1}{ax^{2}+bx+c}}\,\mathrm {d} x={\frac {1}{\sqrt {b^{2}-4ac}}}\ln {\left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|}+C\iff b^{2}-4ac>0} ∫ 1 a x 2 + b x + c d x = 2 4 a c − b 2 arctan ( 2 a x + b 4 a c − b 2 ) + C ⟺ b 2 − 4 a c < 0 {\displaystyle \int {\frac {1}{ax^{2}+bx+c}}\,\mathrm {d} x={\frac {2}{\sqrt {4ac-b^{2}}}}\arctan {\left({\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\right)}+C\iff b^{2}-4ac<0} ∫ x + c ( x + b ) 2 + a 2 d x = 1 2 ln ( x 2 + 2 b x + a 2 + b 2 ) + c − b a arctan ( x + b a ) + C {\displaystyle \int {\frac {x+c}{\left(x+b\right)^{2}+a^{2}}}\,\mathrm {d} x={\frac {1}{2}}\ln {\left(x^{2}+2bx+a^{2}+b^{2}\right)}+{\frac {c-b}{a}}\arctan {\left({\frac {x+b}{a}}\right)}+C} ∫ ln x d x = x ln x − x + C {\displaystyle \int \ln {x}\,\mathrm {d} x=x\ln {x}-x+C} ∫ log b x d x = x log b x − x log b e + C {\displaystyle \int \log _{b}{x}\,\mathrm {d} x=x\log _{b}{x}-x\log _{b}{e}+C} ∫ e x d x = e x + C {\displaystyle \int e^{x}\,\mathrm {d} x=e^{x}+C} ∫ e a x d x = e a x a + C {\displaystyle \int e^{ax}\,\mathrm {d} x={\frac {e^{ax}}{a}}+C} ∫ f ′ ( x ) e f ( x ) d x = e f ( x ) + C {\displaystyle \int f'(x)e^{f(x)}\,\mathrm {d} x=e^{f(x)}+C} ∫ a x d x = a x ln a + C {\displaystyle \int a^{x}\,\mathrm {d} x={\frac {a^{x}}{\ln {a}}}+C} ∫ a f ( x ) f ′ ( x ) d x = a f ( x ) ln a + C {\displaystyle \int a^{f(x)}f'(x)\,\mathrm {d} x={\frac {a^{f(x)}}{\ln {a}}}+C} ∫ 1 1 − x 2 d x = arcsin x + C {\displaystyle \int {1 \over {\sqrt {1-x^{2}}}}\,\mathrm {d} x=\arcsin {x}+C} ∫ − 1 1 − x 2 d x = arccos x + C {\displaystyle \int {-1 \over {\sqrt {1-x^{2}}}}\,\mathrm {d} x=\arccos {x}+C} ∫ 1 | x | x 2 − 1 d x = arcsec x + C {\displaystyle \int {1 \over |x|{\sqrt {x^{2}-1}}}\,\mathrm {d} x=\operatorname {arcsec} {x}+C} ∫ 1 1 + x 2 d x = settsinh x + C {\displaystyle \int {1 \over {\sqrt {1+x^{2}}}}\,\mathrm {d} x=\operatorname {settsinh} {x}+C} ∫ 1 x 2 − 1 d x = settcosh x + C {\displaystyle \int {1 \over {\sqrt {x^{2}-1}}}\,\mathrm {d} x=\operatorname {settcosh} {x}+C} ∫ a 2 − x 2 d x = a 2 2 arcsin x a + x 2 a 2 − x 2 + C {\displaystyle \int {\sqrt[{}]{a^{2}-x^{2}}}\,\mathrm {d} x={\frac {a^{2}}{2}}\arcsin {\frac {x}{a}}+{\frac {x}{2}}\,{\sqrt {a^{2}-x^{2}}}\ +C} ∫ a 2 + x 2 d x = a 2 2 settsinh x a + x 2 a 2 + x 2 + C {\displaystyle \int {\sqrt[{}]{a^{2}+x^{2}}}\,\mathrm {d} x={\frac {a^{2}}{2}}\operatorname {settsinh} {\frac {x}{a}}+{\frac {x}{2}}\,{\sqrt {a^{2}+x^{2}}}\ +C} ∫ cos x d x = sin x + C {\displaystyle \int \cos {x}\,\mathrm {d} x=\sin {x}+C} ∫ sin x d x = − cos x + C {\displaystyle \int \sin {x}\,\mathrm {d} x=-\cos {x}+C} ∫ f ′ ( x ) cos f ( x ) d x = sin f ( x ) + C {\displaystyle \int f'(x)\cos f(x)\,\mathrm {d} x=\sin {f(x)}+C} ∫ f ′ ( x ) sin f ( x ) d x = − cos f ( x ) + C {\displaystyle \int f'(x)\,\sin {f(x)}\,\mathrm {d} x=-\cos {f(x)}+C} ∫ tan x d x = − ln | cos x | + C {\displaystyle \int \tan {x}\,\mathrm {d} x=-\ln {\left|\cos {x}\right|}+C} ∫ csc x d x = − ln | csc x + cot x | + C {\displaystyle \int \csc {x}\,\mathrm {d} x=-\ln {\left|\csc {x}+\cot {x}\right|}+C} ∫ sec x d x = ln | sec x + tan x | + C {\displaystyle \int \sec {x}\,\mathrm {d} x=\ln {\left|\sec {x}+\tan {x}\right|}+C} ∫ cot x d x = ln | sin x | + C {\displaystyle \int \cot {x}\,\mathrm {d} x=\ln {\left|\sin x\right|}+C} ∫ sec 2 x d x = tan x + C {\displaystyle \int \sec ^{2}x\,\mathrm {d} x=\tan x+C} ∫ csc 2 x d x = − cot x + C {\displaystyle \int \csc ^{2}x\,\mathrm {d} x=-\cot x+C} ∫ sin 2 x d x = 1 2 ( x − sin x cos x ) + C {\displaystyle \int \sin ^{2}x\,\mathrm {d} x={\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\,\mathrm {d} x={\frac {1}{2}}(x+\sin x\cos x)+C} ∫ cos ( a x ) d x = 1 a sin ( a x ) + C {\displaystyle \int \cos(ax)\,\mathrm {d} x={\frac {1}{a}}\,\sin(ax)+C} ∫ sin ( a x ) d x = − 1 a cos ( a x ) + C {\displaystyle \int \sin(ax)\,\mathrm {d} x=-{\frac {1}{a}}\cos(ax)+C} ∫ sinh x d x = cosh x + C {\displaystyle \int \sinh x\,\mathrm {d} x=\cosh x+C} ∫ cosh x d x = sinh x + C {\displaystyle \int \cosh x\,\mathrm {d} x=\sinh x+C} ∫ tanh x d x = ln ( cosh x ) + C {\displaystyle \int \tanh x\,\mathrm {d} x=\ln(\cosh x)+C} ∫ csch x d x = ln | tanh x 2 | + C {\displaystyle \int {\mbox{csch}}\,x\,\mathrm {d} x=\ln \left|\tanh {x \over 2}\right|+C} ∫ sech x d x = arctan ( sinh x ) + C {\displaystyle \int {\mbox{sech}}\,x\,\mathrm {d} x=\arctan(\sinh x)+C} ∫ coth x d x = ln | sinh x | + C {\displaystyle \int \coth x\,\mathrm {d} x=\ln |\sinh x|+C} ∫ settcosh x d x = x settcosh x − x 2 − 1 + C {\displaystyle \int \operatorname {settcosh} x\,\mathrm {d} x=x\operatorname {settcosh} x-{\sqrt {x^{2}-1}}+C} ∫ settsinh x d x = x settsinh x − x 2 + 1 + C {\displaystyle \int \operatorname {settsinh} x\,\mathrm {d} x=x\operatorname {settsinh} x-{\sqrt {x^{2}+1}}+C} ∫ setttanh x d x = x setttanh x + log ( 1 − x 2 ) 2 + C {\displaystyle \int \operatorname {setttanh} x\,\mathrm {d} x=x\operatorname {setttanh} x+{\frac {\log {(1-x^{2})}}{2}}+C} 
![{\displaystyle \int [f(x)+g(x)]\,\mathrm {d} x=\int f(x)\,\mathrm {d} x+\int g(x)\,\mathrm {d} x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94d84511b22637e0181a8ed0d52913c323c133ef)
![{\displaystyle \int {\sqrt[{}]{a^{2}-x^{2}}}\,\mathrm {d} x={\frac {a^{2}}{2}}\arcsin {\frac {x}{a}}+{\frac {x}{2}}\,{\sqrt {a^{2}-x^{2}}}\ +C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bad7c4ed71806e5f3c40ca2e3555918b44e2004c)
![{\displaystyle \int {\sqrt[{}]{a^{2}+x^{2}}}\,\mathrm {d} x={\frac {a^{2}}{2}}\operatorname {settsinh} {\frac {x}{a}}+{\frac {x}{2}}\,{\sqrt {a^{2}+x^{2}}}\ +C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/db1acbcc3d5830de81a93d61fd8feb1695ef86af)
