I'm trying to evaluate this integral on $0 \le x \le 1$, but substition and by parts doesn't work here.
$$\int {\frac{\ln(1+x)}{x}dx}$$
By Taylor series, we have: $$\frac{d}{dx}\ln(x+1)=\frac{1}{x+1}=1-x+x^2-x^3+\cdots, |x|<1$$ The integral will be: $$\int_0^{1} {\bigg(1-\frac{x}{2}+\frac{x^2}{3}+\cdots\bigg)dx}=1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\cdots=\frac{\pi^2}{12}$$
Do you know other methods? Thank you.
The integrand has no elementary primitive. If you replace $x$ by $x-1$ you obtain $$\int\frac{\ln x}{x-1}\,dx\ ,$$ which is listed here.
Acknowledgement: thanks to @sas for pointing me towards the above link in response to a question I asked a few days ago.
$\int\dfrac{\ln(1+x)}{x}dx$
$=\int\sum\limits_{n=1}^\infty\dfrac{(-1)^{n+1}x^n}{nx}dx$
$=\int\sum\limits_{n=1}^\infty\dfrac{(-1)^{n+1}x^{n-1}}{n}dx$
$=\sum\limits_{n=1}^\infty\dfrac{(-1)^{n+1}x^n}{n^2}+C$
After substituting $u=-x$ we see that the integral is a dilogarithm: $$\int\frac{\ln (x+1)}{x}\,\mathrm d x=-\int \frac{\ln (1-u)}{u}\,\mathrm d u=-\operatorname{Li}_2(-x)+C$$
