Finding $\int^1_0 \frac{\log(1+x)}{x}dx$ without series expansion I was trying to evaluate  $$\int^1_0 \frac{\log(1+x)}{x}dx.$$
I expanded $\log(1+x) $  as
$x -\frac{x^2}{2}... $ and got the answer. I would like to know if there is any way to do it without series expanding.

With $\int^1_0 \frac{\ln(1+x)}{x}dx
= 2\int^1_0 \frac{\ln(1+x^2)}{x}dx 
= 3\int^1_0 \frac{\ln(1+x^3)}{x}dx $
\begin{align} 
&\int^1_0 \frac{\ln(1+x)}{x}dx\\= &\ \frac67\int^1_0 {\frac{\ln\frac{(1+x^2)(1+x)} {(1+x^3)}}{x} } dx
=\frac67\int^1_0 \frac{\ln\frac{1+x^2}{1-x+x^2}}{x}dx \\= &\ \frac67\int_0^1\int_0^{\frac\pi6}
\frac{2\cos t }{1-2x\sin t + x^2}dt\ dx
=\frac67\int_0^{\frac\pi6}\left(\frac\pi2+t\right)dt
=\frac{\pi^2}{12}
\end{align}
