Popis integrala logaritamskih funkcija

Slijedi popis integrala (antiderivacija funkcija) logaritamskih funkcija. Za potpun popis integrala funkcija, pogledati tablica integrala i popis integrala.

Valja uočiti: u članku se pretpostavlja da vrijedi x>0.

$\int \ln cxdx=x\ln cx-x+C$

$\int \ln (ax+b)dx=x\ln (ax+b)-x+\frac{b}{a}\ln (ax+b)+C$

$\int (\ln x{)}^{2}dx=x(\ln x{)}^2-2x\ln x+2x+C$

$\int (\ln cx{)}^{n}dx=x(\ln cx{)}^n-n\int (\ln cx{)}^{n-1}dx+C$

$\int \frac{dx}{\ln x}=\ln |\ln x|+\ln x+\sum _{i=2}^{\mathrm{\infty }}\frac{(\ln x{)}^i}{i\cdot i!}+C$

$\int \frac{dx}{(\ln x{)}^n}=-\frac{x}{(n-1)(\ln x{)}^{n-1}}+\frac{1}{n-1}\int \frac{dx}{(\ln x{)}^{n-1}}+C\text{(za }n\ne 1\text{)}$

$\int x^m\ln xdx=x^{m+1}(\frac{\ln x}{m+1}-\frac{1}{(m+1{)}^2})+C\text{(za }m\ne -1\text{)}$

$\int x^m(\ln x{)}^{n}dx=\frac{x^{m+1}(\ln x{)}^n}{m+1}-\frac{n}{m+1}\int x^m(\ln x{)}^{n-1}dx+C\text{(za }m\ne -1\text{)}$

$\int \frac{(\ln x{)}^{n}dx}{x}=\frac{(\ln x{)}^{n+1}}{n+1}+C\text{(za }n\ne -1\text{)}$

$\int \frac{\ln x^{n}dx}{x}=\frac{(\ln x^n{)}^2}{2n}+C\text{(za }n\ne 0\text{)}$

$\int \frac{\ln xdx}{x^m}=-\frac{\ln x}{(m-1)x^{m-1}}-\frac{1}{(m-1{)}^2{x}^{m-1}}+C\text{(za }m\ne 1\text{)}$

$\int \frac{(\ln x{)}^{n}dx}{x^m}=-\frac{(\ln x{)}^n}{(m-1)x^{m-1}}+\frac{n}{m-1}\int \frac{(\ln x{)}^{n-1}dx}{x^m}+C\text{(za }m\ne 1\text{)}$

$\int \frac{x^{m}dx}{(\ln x{)}^n}=-\frac{x^{m+1}}{(n-1)(\ln x{)}^{n-1}}+\frac{m+1}{n-1}\int \frac{x^{m}dx}{(\ln x{)}^{n-1}}+C\text{(za }n\ne 1\text{)}$

$\int \frac{dx}{x\ln x}=\ln |\ln x|+C$

$\int \frac{dx}{x^n\ln x}=\ln |\ln x|+\sum _{i=1}^{\mathrm{\infty }}(-1{)}^i\frac{(n-1{)}^i(\ln x{)}^i}{i\cdot i!}+C$

$\int \frac{dx}{x(\ln x{)}^n}=-\frac{1}{(n-1)(\ln x{)}^{n-1}}+C\text{(za }n\ne 1\text{)}$

$\int \ln (x^2+a^2)dx=x\ln (x^2+a^2)-2x+2a{\tan }^{-1}\frac{x}{a}+C$

$\int \frac{x}{x^2+a^2}\ln (x^2+a^2)dx=\frac{1}{4}{\ln }^2(x^2+a^2)+C$

$\int \sin (\ln x)dx=\frac{x}{2}(\sin (\ln x)-\cos (\ln x))+C$

$\int \cos (\ln x)dx=\frac{x}{2}(\sin (\ln x)+\cos (\ln x))+C$

$\int e^x(x\ln x-x-\frac{1}{x})dx=e^x(x\ln x-x-\ln x)+C$

$\int \frac{1}{e^x}(\frac{1}{x}-\ln x)dx=\frac{\ln x}{e^x}+C$

$\int e^x(\frac{1}{\ln x}-\frac{1}{x{\ln }^{2}x})dx=\frac{e^x}{\ln x}+C$

Izvori

• Milton Abramowitz i Irene A. Stegun, Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 1964. Nekolicina integrala je izlistana na stranici 69 ove klasične knjige.