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
Asked
Active
Viewed 54 times
2
-
I suppose that this could involve the polylogarithm function. – Claude Leibovici Jan 06 '17 at 11:26
-
solution is required to become in the parts form – muhammed sulltan Jan 06 '17 at 11:29
-
I have to solve this integral, $$\int\frac{\ln^2(a+bx)}{x^n}dx$$ but I didnt now how to solve. I tried, and during calculation gain that integral, maybe you can help me, to solve the integral $$\int\frac{\ln^2(a+bx)}{x^n}dx$$ – muhammed sulltan Jan 06 '17 at 11:34
1 Answers
0
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.
-
Thanky very much sir, and sorry but I because I did not write well, have: $\int\ln\frac{a+bx}{x}\cdot\frac{dx}{a+bx}$ – muhammed sulltan Jan 06 '17 at 11:48
-
for me with this kind of interest is integral: $$\int\frac{\ln^2(a+bx)}{x^n}$$If possible, please give me some help on this integral – muhammed sulltan Jan 06 '17 at 11:50