Research website of Vyacheslav Gorchilin
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Formula. Math
Integrals containing logarithm
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$\int \ln x\,dx$
$\int \log_a x\,dx$
$\int \ln(ax+b) dx$
$\int x\ln x\,dx$
$\int x\ln(ax+b) dx$
$\int x^2 \ln x\,dx$
$\int x^a \ln x\,dx$
$\int {\ln x \over x} dx$
$\int {\ln x \over x^a} dx$
$\int {\ln(ax+b) \over x^2} dx$
$\int \ln^2 x\,dx$
$\int x\ln^2 x\,dx$
$\int {\ln^2 x \over x}dx$
$\int {\ln^2 x \over x^2}dx$
$\int {dx \over x\ln x}$
$\int {dx \over x(\ln x)^a}$
$\int \ln(x^2+a^2) dx$
$\int x\ln(x^2+a^2) dx$
$\int x^2\ln(x^2+a^2) dx$
$\int \ln|x^2-a^2| dx$
$\int x\ln|x^2-a^2| dx$
$\int \ln(x+\sqrt{x^2+a^2}) dx$
$\int \ln(x+\sqrt{x^2-a^2}) dx$
$\int x\ln(x+\sqrt{x^2+a^2}) dx$
$\int x\ln(x+\sqrt{x^2-a^2}) dx$
$\int \ln x\,dx = x\ln x - x$
$\int \log_a x\,dx = \frac{1}{\ln a}(x\ln x - x), \quad a \neq 1$
$\int \ln(ax+b) dx = \frac{ax+b}{a} \ln(ax+b) - x, \quad a \neq 0$
$\int x\ln x\,dx = \frac{x^2}{2} \ln x - \frac{x^2}{4}$
$\int x\ln(ax+b) dx = {a^2x^2-b^2 \over 2a^2} \ln(ax+b) - {ax^2-2bx \over 4a}, \quad a \neq 0$
$\int x^2 \ln x\,dx = \frac{x^3}{3} \ln x - \frac{x^3}{9}$
$\int x^a \ln x\,dx = \frac{x^{a+1}}{a+1} \ln x - \frac{x^{a+1}}{(a+1)^2}, \quad a \neq 1$
$\int {\ln x \over x} dx = {(\ln x)^2 \over 2}$
$\int {\ln x \over x^a} dx = {\ln x \over (1-a)\,x^{a-1}} - {1 \over (1-a)^2x^{a-1}}, \quad a \neq 1$
$\int {\ln(ax+b) \over x^2} dx = \frac{a}{b} \ln x - {ax+b \over bx} \ln(ax+b), \quad b \neq 0$
$\int \ln^2 x\,dx = x\ln^2 x - 2x\ln x + 2x$
$\int x\ln^2 x\,dx = \frac{x^2}{2}\ln^2 x - \frac{x^2}{2}\ln x + \frac{x^2}{4}$
$\int {\ln^2 x \over x}dx = \frac13\ln^3 x$
$\int {\ln^2 x \over x^2}dx = - {\ln^2 x \over x} - {2\ln x \over x} - \frac{2}{x}$
$\int {dx \over x\ln x} = \ln|\ln x|$
$\int {dx \over x(\ln x)^a} = {1 \over (1-a)(\ln x)^{a-1}}, \quad a \neq 1$
$\int \ln(x^2+a^2) dx = x\ln(x^2+a^2) - 2x + 2a\,\mathtt{arctg}(\frac{x}{a})$
$\int x\ln(x^2+a^2) dx = {x^2+a^2 \over 2}\ln(x^2+a^2) - \frac{x^2}{2}$
$\int x^2\ln(x^2+a^2) dx = {x^3 \over 3}\ln(x^2+a^2) - \frac{2x^3}{9} + \frac{2a^2x}{3} - \frac{2a^3}{3}\mathtt{arctg}(\frac{x}{a})$
$\int \ln|x^2-a^2| dx = x\ln|x^2-a^2| - 2x - a\ln\left|{x+a \over x-a}\right|$
$\int x\ln|x^2-a^2| dx = {x^2-a^2 \over 2}\ln|x^2-a^2| - \frac{x^2}{2}$
$\int \ln(x+\sqrt{x^2+a^2}) dx = x\ln(x+\sqrt{x^2+a^2}) - \sqrt{x^2+a^2}$
$\int \ln(x+\sqrt{x^2-a^2}) dx = x\ln(x+\sqrt{x^2-a^2}) - \sqrt{x^2-a^2}$
$\int x\ln(x+\sqrt{x^2+a^2}) dx = {2x^2+a^2 \over 4}\ln(x+\sqrt{x^2+a^2}) - {x\sqrt{x^2+a^2} \over 4}$
$\int x\ln(x+\sqrt{x^2-a^2}) dx = {2x^2-a^2 \over 4}\ln(x+\sqrt{x^2-a^2}) - {x\sqrt{x^2-a^2} \over 4}$