Finding $ \int^1_0 \frac{\ln(1+x)}{x}dx$ There is supposed to be a clean solution to the integral below, maybe involving some symmetry
$$ \int^1_0 \frac{\ln(1+x)}{x}dx$$
I have tried integration by parts as followed:
$\ln(x+1)=u$ ,$\frac{1}{x+1} dx = du$ and also $\frac{1}{x}dx=dv$, $\ln(x)=v$. Then, the integral becomes
$$\int^1_0 \frac{\ln(1+x)}{x}dx=- \int^1_0 \frac{\ln x}{1+x}dx$$
which does not make this easier.
I have also tried using the identity $\int_a^bf(x)~dx=\int_a^bf(a+b-x)~dx$. So let $I = \int^1_0 \frac{\ln(1+x)}{x}dx$ and also $I = \int^1_0 \frac{\ln(2-x)}{1-x}  $. Then
$$2I= \int^1_0 \ln(1+x)+\ln(2-x)$$
which is not any easier, either.
Any ideas? :)
 A: Hint. One may recall that
$$
\ln (1+x)=\sum_{n=1}^\infty(-1)^{n-1}\frac{x^n}{n},\quad |x|<1,
$$ one may then divide by $x$ and one is allowed to integrate termwise obtaining
$$
\int_0^1\frac{\ln (1+x)}x\:dx=\sum_{n=1}^\infty(-1)^{n-1}\int_0^1\frac{x^{n-1}}{n}=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}.
$$ Can you take it from here?
A: By integration by parts:
$$ \int_{0}^{1}\frac{\log(1+x)\,dx}{x} = -\int_{0}^{1}\frac{\log x}{1+x}\,dx=2\int_{0}^{1}\frac{-\log x}{1-x^2}\,dx-\int_{0}^{1}\frac{-\log x}{1-x} $$
but since $\int_{0}^{1}(-\log x)x^k\,dx = \frac{1}{(k+1)^2}$, by expanding $\frac{1}{1-x^2}$ and $\frac{1}{1-x}$ as geometric series we get:
$$ \int_{0}^{1}\frac{\log(1+x)\,dx}{x} = 2\sum_{k\text{ odd}}\frac{1}{k^2}-\sum_{k\geq 1}\frac{1}{k^2}=\sum_{k\geq 1}\frac{1}{k^2}-2\sum_{k\text{ even}}\frac{1}{k^2}=\frac{1}{2}\sum_{k\geq 1}\frac{1}{k^2}=\color{blue}{\frac{\pi^2}{12}}. $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
\int_{0}^{1}{\ln\pars{1 + x} \over x}\,\dd x &
\,\,\,\stackrel{x\ \mapsto\ -x}{=}\,\,\,
\int_{0}^{-1}{\ln\pars{1 - x} \over x}\,\dd x =
-\int_{0}^{-1}\mrm{Li}_{2}'\pars{x}\,\dd x =
-\mrm{Li}_{2}\pars{-1} + \mrm{Li}_{2}\pars{0}
\\[5mm] & = \bbx{\ds{\pi^{2} \over 12}} \\ &
\end{align}
A: If you know the gamma function $\Gamma$, the Dirichlet function $\eta$ and the formula 
$$F(n):=\int_0^{\infty}\frac{u^{n-1}}{e^u+1}\; \mathrm du=\Gamma(n)\cdot\eta(n) \qquad (n\in \mathbf N)$$
you can integrate
$$I:= \int_0^{1}\frac{\ln(1+x)}{x} \; \mathrm d x$$
by parts to get
$$I=-\int^1_0 \frac{\log(x)}{x+1} \; \mathrm d x.$$
If you substitue $x=\mathrm e^u$ ($\mathrm d x = \mathrm e^u \mathrm du$) you will get
$$ \int_0^{\infty}\frac{u}{e^u+1}\mathrm du.$$ Therefore you get
$$F(2)=I\stackrel{.}{=}0.82246703342$$
(which is equal to $\frac{\pi^2}{12}$).
A: HINT:
use the expansion $$ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots$$
$$\frac{ln(1+x)}{x}=1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\ldots$$
$$\int \frac{ln(1+x)}{x}\ dx=x-\frac{x^2}{2^2}+\frac{x^3}{3^2}-\frac{x^4}{4^2}+\ldots$$
A: Here is an elementary integration of $I=\int^1_0 \frac{\ln(1+x)}{x}dx $
\begin{align}
I=-\frac12 \int^1_0 \underset{x\to x^3}{\frac{\ln(1+x)}{x}dx }+\frac32\int^1_0 \frac{\ln(1+x)}{x}dx 
=-\frac32\int^1_0 \frac{\ln(1-x+x^2)}{x}dx 
\end{align}
Let $J(a)=\int^1_0 \frac{\ln(1-2x\sin a+x^2)}{x}dx$
$$J’(a)=-\int^1_0 \frac{2\cos a\ }{(x-\sin a)^2 +\cos^2a}dx
=-\left(\frac\pi2+a\right)$$
Then, with
$J(0)=\int^1_0 {\frac{\ln(1+x^2)}{x}dx } \overset{x^2\to x} =\frac12I $
$$I= -\frac32J(\frac\pi6)=  -\frac32\left( J(0)+\int_0^{\frac\pi6}J’(a)da\right)=-\frac34I +\frac32\int_0^{\frac\pi6}\left(\frac\pi2+a\right)da
$$
which leads to
$$I= \frac{\pi^2}{12}$$
A: $$
\begin{aligned}
\textrm{Using the series of }
\ln (1+y) &=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+1} y^{n+1}, \textrm{ we have } \\
\int_{0}^{1} \frac{\ln (1+y)}{y} d y &=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+1} \int_{0}^{1} y^{n} d y \\
&=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(n+1)^{2}} \\
&=\zeta(2)-2 \sum_{n=0}^{\infty} \frac{1}{(2 n)^{2}} \\
&=\frac{1}{2} \zeta(2) \\
&=\frac{\pi^{2}}{12}
\end{aligned}
$$
