How to solve the given integral

Question

I didnt now how to solve the integral: $$\int\frac{\frac{\ln(a+bx)}{x}}{a+bx}dx$$ Please help me. Thanks for your help

Answer 1

We have $$I =\int \frac {\ln (a+bx)}{x (ax+b)} dx =\frac {1}{a}\int \frac {\ln (a+bx)}{x} dx -\frac {b}{a}\int \frac {\ln (a+bx)}{a+bx} dx =\frac {1}{a}I_1 -\frac {b}{a}I_2$$

$I_1$ can be solved as: $$I_1 =\int \frac {\ln (\frac {bx}{a}+1)}{x} dx +\ln a \int \frac {1}{x} dx $$ Substituting $u=-\frac {bx}{a} $, we get $-\int -\frac {\ln (1-u)}{u} du =-\text {Li}_{2}(u) $ where $\text {Li}_{2}(u) $ is the dilogarithm.

$I_2$ is easy to solve by substituting $v =\ln (a+bx) $. Hope it helps.