# Matematika/Integrálszámítás/Szabályok

Ugrás a navigációhoz Ugrás a kereséshez

## Alapintegrálok

Alapintegráloknak nevezzük az elemi valós függvények differenciálási szabályainak megfordításából adódó primitív függvényeket.

 ${\displaystyle \int x^{n}\,dx}$ ${\displaystyle ={\frac {x^{n+1}}{n+1}}+c}$ ${\displaystyle (x\in \mathbb {R} ,n\in \mathbb {N} )}$ ${\displaystyle \int x^{\alpha }\,dx}$ ${\displaystyle ={\frac {x^{\alpha +1}}{\alpha +1}}+c}$ ${\displaystyle (x\in \mathbb {R} ^{+},-1\neq \alpha \in \mathbb {R} )}$ ${\displaystyle \int {\frac {1}{x}}\,dx}$ ${\displaystyle =\,\ln |x|+c}$ ${\displaystyle (0\neq x\in \mathbb {R} )}$ ${\displaystyle \int e^{x}\,dx}$ ${\displaystyle =\,e^{x}+c}$ ${\displaystyle (x\in \mathbb {R} )}$ ${\displaystyle \int a^{x}\,dx}$ ${\displaystyle ={\frac {a^{x}}{\ln a}}+c}$ ${\displaystyle (x\in \mathbb {R} ,1\neq a\in \mathbb {R} ^{+})}$ ${\displaystyle \int \sin x\,dx}$ ${\displaystyle =-\cos x\,+c}$ ${\displaystyle (x\in \mathbb {R} )}$ ${\displaystyle \int \cos x\,dx}$ ${\displaystyle =\sin x\,+c}$ ${\displaystyle (x\in \mathbb {R} )}$ ${\displaystyle \int {\frac {1}{\sin ^{2}x}}\,dx}$ ${\displaystyle =-\mathrm {ctg} \,x\,+c}$ ${\displaystyle (k\pi \neq x\in \mathbb {R} ,k\in \mathbb {Z} )}$ ${\displaystyle \int {\frac {1}{\cos ^{2}x}}\,dx}$ ${\displaystyle =\mathrm {tg} \,x\,+c}$ ${\displaystyle \left({\frac {k\pi }{2}}\neq x\in \mathbb {R} ,k\in \mathbb {Z} \right)}$ ${\displaystyle \int \mathrm {sh} \,x\,dx}$ ${\displaystyle =\mathrm {ch} \,x\,+c}$ ${\displaystyle (x\in \mathbb {R} )}$ ${\displaystyle \int \mathrm {ch} \,x\,dx}$ ${\displaystyle =\mathrm {sh} \,x\,+c}$ ${\displaystyle (x\in \mathbb {R} )}$ ${\displaystyle \int {\frac {1}{\mathrm {sh} ^{2}x}}\,dx}$ ${\displaystyle =-\mathrm {cth} \,x\,+c}$ ${\displaystyle (0\neq x\in \mathbb {R} )}$ ${\displaystyle \int {\frac {1}{\mathrm {ch} ^{2}x}}\,dx}$ ${\displaystyle =\mathrm {th} \,x\,+c}$ ${\displaystyle (x\in \mathbb {R} )}$ ${\displaystyle \int {\frac {1}{1+x^{2}}}\,dx}$ ${\displaystyle =\mathrm {arc\,tg} \,x\,+c}$ ${\displaystyle (x\in \mathbb {R} )}$ ${\displaystyle \int {\frac {1}{1-x^{2}}}\,dx}$ ${\displaystyle ={\frac {1}{2}}\ln \left|{\frac {x+1}{x-1}}\right|+c}$ ${\displaystyle =\left\{{\mathrm {ar\,th} \,x+c\quad (1>|x|\in \mathbb {R} ) \atop \mathrm {ar\,ch} \,x+c\quad (1<|x|\in \mathbb {R} )}\right.}$ ${\displaystyle \int {\frac {1}{\sqrt {1-x^{2}}}}\,dx}$ ${\displaystyle =\mathrm {arc\,sin} x\,+c}$ ${\displaystyle (1>|x|\in \mathbb {R} )}$ ${\displaystyle \int {\frac {1}{\sqrt {1+x^{2}}}}\,dx}$ ${\displaystyle =\mathrm {ar\,sh} \,x+c}$ ${\displaystyle (x\in \mathbb {R} )}$ ${\displaystyle \int {\frac {1}{\sqrt {x^{2}-1}}}\,dx}$ ${\displaystyle =\ln \left|x+{\sqrt {x^{2}-1}}\right|+c}$ ${\displaystyle =\left\{\;{\mathrm {ar\,ch} \,x+c\quad \quad (1x\in \mathbb {R} )}\right.}$