TITLE: Finding $\int^1_0 \frac{\log(1+x)}{x}dx$ without series expansion
QUESTION [15 upvotes]: 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.

REPLY [2 votes]: Note that the value of $\int^1_0 \frac{n\ln(1+x^n)}{x}dx$ is invariant of $n$. Then
\begin{align} 
\int^1_0 \frac{\ln(1+x)}{x}dx= -\frac67\int^1_0 {\frac{\ln\frac{(1+x^3)}{(1+x^2)(1+x)}}{x} } dx
=-\frac67\int^1_0 \frac{\ln\left( 1-\frac x{1+x^2}\right)}{x}dx \\
\end{align}
Let
$J(t)=\int^1_0 \frac{\ln\left( 1-\frac {2x\sin t}{1+x^2}\right)}{x}dx$ and
$$J’(t)=-\int^1_0 \frac{2\cos t\ dx }{(x-\sin t)^2 +\cos^2t}=-\left(\frac\pi2+t\right)$$
Thus
$$\int^1_0 \frac{\ln(1+x)}{x}dx = -\frac67 J(\frac\pi6)=  -\frac67\int_0^{\frac\pi6}J’(t)dt=\frac67\int_0^{\frac\pi6}\left(\frac\pi2+t\right)dt
= \frac{\pi^2}{12}$$