Tabela integralov

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Integriranje je ena od dveh osnovnih operacij v infinitezimalnem računu. Ker za razliko od odvajanja ni trivialna, nam včasih pridejo prav tabele znanih integralov. Ta stran navaja nekaj najbolj znanih integralov.

Nedoločeni integrali[uredi | uredi kodo]

Za integracijsko konstanto uporabljamo oznako C in jo lahko določimo, če je znana vrednost primitivne funkcije v neki točki. V splošnem pa je konstanta C nedoločena.

Potence, koreni[uredi | uredi kodo]

\int x^n\,dx =  \frac{x^{n+1}}{n+1} + C\qquad\mbox{ pri }n \ne -1
\int x^{-1}\,dx =\int\frac{dx}{x}= \ln{\left|x\right|} + C
\int \sqrt{x}\,dx=\frac 2 3 x^{3/2} + C \!\,
\int \frac {1}{\sqrt{x}}\,dx=2\sqrt{x} + C \!\,
\int {1 \over \sqrt{1-x^2}} \, dx = \arcsin {x} + C
\int {x \over \sqrt{x^2-1}} \, dx = \mbox{arcsec}\,{x} + C

Polinomi, racionalne funkcije[uredi | uredi kodo]

\int (ax+b)\,dx=\frac{ax^{2}}{2}+bx + C \!\,
\int (a x^{2} + bx + c)\,dx=\frac{a}{3}x^{3}+\frac{b}{2}x^{2} + cx + C \!\,
\int (ax+b)^{n}\,dx=\frac{(ax+b)^{n+1}}{a(n+1)} + C \!\,
\int \frac{dx}{ax+b}=\frac{1}{a} \ln|ax+b| + C \!\,
\int \frac{1}{x^2+1} \, dx = \arctan{x} + C
\int \frac{dx}{x^2+a^2}=\frac{1}{a} \arctan\frac{x}{a} + C \!\,
\int \frac{f'(x)}{f(x)}\,dx=\ln |f(x)| +C

Eksponentne, logaritemske funkcije[uredi | uredi kodo]

\int e^x\,dx = e^x + C
\int e^{cx}\;\mathrm{d}x = \frac{1}{c} e^{cx}+ C
\int a^x\,dx = \frac{a^x}{\ln{a}} + C
\int x e^{x}\,dx=e^{x}(x-1) + C \!\,
\int \frac{dx}{e^{x}}=-\frac{1}{e^{x}} + C \!\,
\int \frac{x}{e^{x}}\,dx=-\frac{x+1}{e^{x}} + C \!\,
\int \frac{e^{x}}{x}\,dx=-\operatorname{Ei}(-x) + C \!\,      Opomba: Ei = eksponentni integral
\int \ln {x}\,dx = x \ln {x} - x + C
\int \log_a x\,dx= x \log_a {x} - \frac{x}{\ln a} + C

Trigonometrične funkcije[uredi | uredi kodo]

\int \cos{x}\, dx = \sin{x} + C
\int \cos({nx}) \, dx = {sin (nx) \over n} + C
\int \sin{x}\, dx = -\cos{x} + C
\int \sin({nx}) \, dx = {-{cos(nx) \over n}} + C
\int \tan{x} \, dx = -\ln{\left| \cos {x} \right|} + C
\int \csc{x} \, dx = -\ln{\left| \csc{x} + \cot{x}\right|} + C
\int \sec{x} \, dx = \ln{\left| \sec{x} + \tan{x}\right|} + C
\int \cot{x} \, dx = \ln{\left| \sin{x} \right|} + C
\int \frac{dx}{\cos^2 x}=\int \sec^2 x \, dx = \tan x + C
\int \frac{dx}{\sin^2 x}=\int \csc^2 x \, dx = -\cot x + C
\int \sin^2 x \, dx = {2x - \sin 2x \over 4} + C = {{x\over 2} - {sin 2x \over 4}} + C
\int \cos^2 x \, dx = {2x + \sin 2x \over 4} + C = {{x\over 2} + {sin 2x \over 4}} + C

Hiperbolične funkcije[uredi | uredi kodo]

\int \sinh x \, dx = \cosh x + C
\int \cosh x \, dx = \sinh x + C
\int \tanh x \, dx = \ln |\cosh x| + C
\int \coth x \, dx = \ln|\sinh x| + C
\int \mbox{csch}\,x \, dx = \ln\left| \tanh {x \over2}\right| + C
\int \mbox{sech}\,x \, dx = \arctan(\sinh x) + C

Določeni integrali[uredi | uredi kodo]

Obstajajo funkcije katerih primitivnih funkcij ne moremo izraziti v zaprti obliki. Vendar lahko izračunamo vrednosti določenih integralov teh funkcij v nekaterih intervalih. Nekaj uporabnih določenih integralov je podanih spodaj.

\int_{0}^{\infty} { \frac{1}{x^{2}+a^{2}}\,dx} = \frac{\pi}{2a} \!\,
\int_{0}^{\infty} { \frac{1}{(1+x)x^{a}}\,dx} = \frac{\pi}{\sin (a \pi)}, \quad (a < 1) \!\,
\int_{0}^{\infty} { \frac{x^{a-1}}{1+x}\,dx} = \frac{\pi}{\sin (a \pi)}, \quad (0 < a < 1) \!\,
\int_{0}^{\infty} { \frac{1}{(1-x)x^{a}}\,dx} = -\pi \operatorname{ctg}, \quad (a < 1) \!\,
\int_{-\infty}^{\infty} \frac{1}{(1 + x^2/a)^{(a + 1)/2}}\,dx = \frac { \sqrt{a \pi} \ \Gamma(a/2)} {\Gamma \Big[(a + 1)/2) \Big]}, \quad (a > 0) \!\,      (povezava z gostoto verjetnosti Studentove t-porazdelitve)
\int_{0}^{\infty} { \frac{x^{a-1}}{1+x^{b}}\,dx} = \frac{\pi}{b \sin(a\pi/b)}, \quad (0 < a < b) \!\,
\int_{0}^{\infty} { \frac{x^{a}}{x^{b}+c^{b}}\,dx} = \frac{\pi c^{a+1-b}}{b \sin \Big[(a+1)\pi/b \Big]}, \quad (0 < a+1 < b) \!\,
\int_{0}^{\infty} { \frac{x^{a}}{(x^{b}+c^{b})^{d}}\,dx} = \frac{(-1)^{d-1} \pi c^{a+1-bd}}{b \sin \Big[(a+1)\pi/b \Big] (d-1)! \, \Gamma \Big[(a+1)/(b-d+1)\Big]}, \quad (0 < a+1 < bd) \!\,
\int_{0}^{\infty} { \frac{1}{1+2x \, \cos (a) + x^{2}}\,dx} = \frac{a}{\sin a} \!\,
\int_{0}^{\infty} { \frac{x^{b}}{1+2x \, \cos (a) + x^{2}}\,dx} =  \frac{\pi}{\sin (b\pi)} \frac{\sin(ab)}{\sin a}, \quad \left(0 < a < \frac{\pi}{2}\right) \!\,


\int_{0}^{\infty} { \frac{1}{e^{a x}}\,dx} = \frac{1}{a}, \quad (a > 0) \!\,
\int_{-\infty}^{\infty} {\frac{1}{e^{a x^{2}}}\,dx} = 2 \int_{0}^{\infty} {\frac{1}{e^{a x^{2}}}\,dx} = \sqrt{\frac{\pi}{a}}, \quad (a > 0) \!\,      (Gaussov integral)
\int_{0}^{\infty} { \frac{1}{e^{a^{2} x^{2}}}\,dx} = \frac{\sqrt{\pi}}{2a}, \quad (a > 0) \!\,
\int_{0}^{\infty} { \frac{1}{e^{x^{2}+(a^{2} / x^{2})}}\,dx} = \frac{\sqrt{\pi}}{2e^{2a}} \!\,
\int_{-\infty}^{\infty} { \frac{1}{e^{ax^{2} + 2bx}}\,dx} = \sqrt{\frac{\pi}{a}} e^{b^{2}/a}, \quad (a > 0) \!\,
\int_{-\infty}^{\infty} { \frac{1}{e^{ax^{2} + bx + c}}\,dx} = \frac{\sqrt{\pi}}{a} e^{(b^{2}-4ac)/4a}, \quad (a > 0) \!\,
\int_{0}^{\infty} { \frac{x}{e^{x^{2}}}\,dx} = \frac{1}{2} \!\,
\int_{0}^{\infty} { \frac{x}{e^{a (x-b)^{2}}}\,dx} = b \sqrt{\frac{\pi}{a}}, \quad (a > 0) \!\,
\int_{0}^{\infty} { \frac{x^{2}}{e^{a x^{2}}}\,dx} = \frac{1}{2}\sqrt{ \frac{\pi}{a^{3}} }, \quad (a > 0) \!\,
\int_{0}^{\infty} { \frac{x^{2}}{e^{a^{2} x^{2}}}\,dx} = \frac{\sqrt{\pi}}{4a^{3}}, \quad (a > 0) \!\,
\int_{0}^{\infty} { \frac{x^{n}}{e^{a x}}\,dx} = 
\begin{cases}
   \frac{\Gamma(n+1)}{a^{n+1}}, & \ (a > 0, n > -1) \\
   \frac{n!}{a^{n+1}},          & \ (a > 0, n \ge 0; \, (n \in \N_{0}))
\end{cases} \!\,
\int_{0}^{\infty} { \frac{x^{n}}{e^{a x^{2}}}\,dx} = 
\begin{cases}
   \frac{\Gamma \big[(n+1)/2 \big]}{2a^{(n+1)/2}}, & \ (a > 0, n > -1) \\
   \frac{(2k-1)!!}{2^{k+1} a^{k}},       & \ (a > 0, n = 2k, k \in \Z) \\
   \frac{k!}{2a^{k+1}},                  & \ (a > 0, n = 2k + 1, k \in \Z)
\end{cases} \!\,      (!! je dvojna fakulteta)
\int_{0}^{\infty} { \frac{x^{2n}}{e^{ax^{2}}}\,dx} = \frac{1\cdot 3\cdot 5 \cdots (2n-1)}{2^{n+1}a^{n}}\sqrt{\frac{\pi}{a}} \!\,
\int_{0}^{\infty} { \frac{x^{2n+1}}{e^{ax^{2}}}\,dx} = \frac{n!}{2a^{n+1}} , \quad (a > 0, n > -1) \!\,
\int_{0}^{\infty}  \frac{x^{n}}{ne^{x}}\,dx = \Gamma(n) \!\,      (funkcija Γ)


\int_{0}^{\infty} \frac{\ln x}{e^{x}}\,dx = - \gamma = -0,5772156649\ldots \!\,      (Euler-Mascheronijeva konstanta)
\int_{0}^{\infty} { \frac{\sqrt{x}}{e^{a x}}\,dx} = \frac{1}{2a}\sqrt {\frac{\pi}{a}} \!\,
\int_{0}^{\infty} {\frac{1}{e^{x} \sqrt{x}}\,dx} = \Gamma \left( \frac{1}{2} \right) = \sqrt{\pi} \!\,      (Eulerjev integral)
\int_{0}^{\infty} {\frac{1}{e^{a x} \sqrt{x}}\,dx} = \sqrt{\frac{\pi}{a}} \!\,
\int_{0}^{\infty} {\frac{x}{e^x-1}\,dx} = \Gamma (2) \zeta (2) = \frac{\pi^2}{6}      (\zeta(\cdot) jeRiemannova funkcija ζ (baselski problem))
\int_{0}^{\infty} {\frac{x^3}{e^x-1}\,dx} = \Gamma (4) \zeta (4) = \frac{\pi^4}{15}
\int_{0}^{\infty} {\frac{x^{n}}{e^{x}-1}\,dx} = \Gamma (n+1) \zeta (n+1)
\int_{0}^{\infty} \frac{x \ln x}{e^{2\pi x} - 1}\,dx = \frac{1}{24} - \frac{1}{2} \ln A \!\,      (A je Glaisher-Kinkelinova konstanta)
\int_{0}^{1/2} \ln \Gamma(x + 1) \,dx = -\frac{1}{2} - \frac{7}{24} \ln 2 + \frac{1}{4} \ln \pi + \frac{3}{2} \ln A \!\,

Glej tudi[uredi | uredi kodo]