$\int_0^1\frac{\ln{x}\ln{(1+x)}}{1+x}dx$ I want to solve for the following Integral:
$$\int_0^1\frac{\ln{x}\ln{(1+x)}}{1+x}dx$$
I have tried to use:
$$\ln{(1+x)}=-\sum_{k=1}^\infty\frac{(-1)^kx^k}{k}$$
and so
$$\int_0^1\frac{\ln{x}\ln{(1+x)}}{1+x}dx=-\sum_{k=1}^\infty\frac{(-1)^k}{k}\int_0^1\frac{x^k\ln{x}}{1+x}dx$$
 A: The integral is $\frac 1 2\int_0^{1} f(g^{2})'(x)dx$ where $g(x)=\log (1+x)$. Integrating by parts we get $\frac 1 2 [fg^{2}|_0^{1}-\int_0^{1} \frac {g(x)^{2}} x dx]$. To compute the integral in the second term make the substitution $y=\log (1+x)$. You will now get something familiar and I will let you complete the evaluation. 
A: Let,
\begin{align}
U&=\int_0^1\frac{\ln(1+x)\ln x}{1+x}\,dx\\
W&=\int_0^1\frac{\ln^2\left(\frac{x}{1+x}\right)}{1+x}\,dx\\
\end{align} Perform the change of variable $y=\dfrac{x}{1+x}$
\begin{align}
W&=\int_0^{\frac{1}{2}}\frac{\ln^2 x}{1-x}\,dx\\
&=\int_0^1\frac{\ln^2 x}{1-x}\,dx-\int_{\frac{1}{2}}^1\frac{\ln^2 x}{1-x}\,dx\\
\end{align} In the latter integral perform the change of variable $y=\dfrac{1-x}{x}$
\begin{align}
W&=\int_0^1\frac{\ln^2 x}{1-x}\,dx-\int_0^1\frac{\ln^2(1+x)}{x(1+x)}\,dx\\
&=\int_0^1\frac{\ln^2 x}{1-x}\,dx+\int_0^1\frac{\ln^2(1+x)}{1+x}\,dx-\int_0^1\frac{\ln^2(1+x)}{x}\,dx\\
&=\int_0^1\frac{\ln^2 x}{1-x}\,dx+\int_0^1\frac{\ln^2(1+x)}{1+x}\,dx-\Big[\ln x\ln^2(1+x)\Big]_0^1+2\int_0^1 \frac{\ln x\ln(1+x)}{1+x}\,dx\\ 
&=\int_0^1\frac{\ln^2 x}{1-x}\,dx+\int_0^1\frac{\ln^2(1+x)}{1+x}\,dx+2U\\
\end{align} On the other hand,
\begin{align}
W&=\int_0^1\frac{\left(\ln x-\ln(1+x)\right)^2}{1+x}\,dx\\
&=\int_0^1\frac{\ln^2 x}{1+x}\,dx+\int_0^1\frac{\ln^2(1+x)}{1+x}\,dx-2U\\
\end{align} Therefore,
\begin{align}
U&=\frac{1}{4}\left(\int_0^1\frac{\ln^2 x}{1+x}\,dx-\int_0^1\frac{\ln^2 x}{1-x}\,dx\right)\\
&=-\frac{1}{4}\int_0^1\frac{2x\ln^2 x}{1-x^2}\,dx
\end{align} Perform the change of variable $y=x^2$,
\begin{align}
U&=-\frac{1}{16}\int_0^1\frac{\ln^2 x}{1-x}\,dx\\
&=-\frac{1}{16}\times 2\zeta(3)\\
&=\boxed{-\frac{1}{8}\zeta(3)}\\
\end{align}
NB:
I assume only that,
\begin{align}\int_0^1\frac{\ln^2 x}{1-x}\,dx=2\zeta(3)\end{align}
$\displaystyle \left(\dfrac{\ln^2 x}{1-x}=\sum_{k=0}^\infty x^k\ln^2 x,x\in ]0;1]\right)$
A: Hint: without power series: use the substitution $t=\ln(1+x)$.
A: Let
$$I = \int_0^1 \frac{\ln x \ln (1 + x)}{1 + x} \, dx.$$
Integrating by parts we have
$$I = - \frac{1}{2} \int_0^1 \frac{\ln^2 (1 + x)}{x} \, dx \tag1$$
The integral appearing in (1) can be calculated in various ways. One way I have already shown here. The result is $\zeta (3)/4$ where $\zeta (z)$ is the Riemann zeta function. Thus
$$\int_0^1 \frac{\ln x \ln (1 + x)}{1 + x} \, dx = -\frac{1}{8} \zeta (3).$$
