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 $$\int_0^1\frac{\ln{x}\ln{(1+x)}}{1+x}dx=-\sum_{k=1}^\infty\frac{(-1)^k}{k}\int_0^1\frac{x^k\ln{x}}{1+x}dx$$
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.
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}
$\displaystyle \left(\dfrac{\ln^2 x}{1-x}=\sum_{k=0}^\infty x^k\ln^2 x,x\in ]0;1]\right)$
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 $$\int_0^1 \frac{\ln x \ln (1 + x)}{1 + x} \, dx = -\frac{1}{8} \zeta (3).$$
I shall begin with this integral. Recall the generating function:
$$\sum_{n=1}^{\infty} \mathcal{H}_n x^n =-\frac{\ln (1-x)}{1-x} \tag{1}$$
Now setting $x \mapsto -x$ back at $(1)$ we have that:
$$\sum_{n=1}^{\infty} (-1)^n \mathcal{H}_n x^n =- \frac{\ln (1+x)}{1+x} \tag{2}$$
Thus
\begin{align*} \mathcal{K} &= \int_{0}^{1} \frac{\ln x \ln (1+x)}{1+x} \, {\rm d}x \\ &\overset{(2)}{=}-\int_{0}^{1} \ln x \sum_{n=1}^{\infty} (-1)^n \mathcal{H}_n x^n \, {\rm d}x \\ &= - \sum_{n=1}^{\infty} (-1)^n \mathcal{H}_n \int_{0}^{1} x^n \ln x \, {\rm d}x \\ &= \sum_{n=1}^{\infty} \frac{(-1)^n \mathcal{H}_n}{(n+1)^2} \\ &= - \sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_{n-1}}{n^2} \\ &= - \sum_{n=2}^{\infty} \frac{(-1)^n \left [ \mathcal{H}_n - \frac{1}{n} \right ]}{n^2} \\ &= -\sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_n}{n^2} +\sum_{n=2}^{\infty} \frac{(-1)^n}{n^3} \\ &= - \sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_n}{n^2} +1 - \frac{3\zeta(3)}{4} \\ &=- \sum_{n=1}^{\infty} \frac{(-1)^n \mathcal{H}_n}{n^2} - \frac{3\zeta(3)}{4} \\ &= \frac{5 \zeta(3)}{8} - \frac{3\zeta(3)}{4} \\ &= - \frac{\zeta(3)}{8} \end{align*}
$\displaystyle K=\int_0^1 \dfrac{\ln x\ln(1+x)}{1+x}dx=\Big[S(x)\ln (1+x)\Big]_0^1-\int_0^1 \dfrac{S(x)}{1+x}dx$
Define the function $S$ for all $x\in[0;1]$,
$\displaystyle S(x)=\int_0^x \dfrac{\ln t}{1+t}dt=\int_0^1 \dfrac{x\ln(xt)}{1+tx}dt$
Observe that $S$ is continuous on $[0;1]$ and $S(1)=-\dfrac{\pi^2}{12}$.
$\displaystyle K=-\dfrac{1}{12}\pi^2\ln 2-\int_0^1\int_0^1 \dfrac{x\ln(xt)}{(1+tx)(1+x)}dtdx$
$\displaystyle K=-\dfrac{1}{12}\pi^2\ln 2-\int_0^1\int_0^1 \dfrac{x\ln(x)}{(1+tx)(1+x)}dtdx-\int_0^1\int_0^1 \dfrac{x\ln(t)}{(1+tx)(1+x)}dtdx$
$\displaystyle K=-\dfrac{1}{12}\pi^2\ln 2-\int_0^1 \left[\dfrac{\ln x\ln(1+tx)}{1+x}\right]_{t=0}^{t=1}dx-\int_0^1 \left[\dfrac{\ln t\ln(1+x)-\ln t\ln(1+tx)}{t-1}+\dfrac{\ln t\ln(1+tx)}{t}\right]_{x=0}^{x=1}dt$
$\displaystyle K=-\dfrac{1}{12}\pi^2\ln 2-K+\ln 2\int_0^1 \dfrac{\ln t}{1-t}dt-J-\int_0^1 \dfrac{\ln t\ln(1+t)}{t}dt$
$\displaystyle 2K=-\dfrac{1}{12}\pi^2\ln 2-\dfrac{1}{6}\pi^2\ln 2+\dfrac{1}{4}\pi^2\ln 2-\zeta(3)+\dfrac{3}{4}\zeta(3)$
$\boxed{K=\displaystyle -\dfrac{1}{8}\zeta(3)}$
$K=\displaystyle \int_0^1 \dfrac{\ln x\ln(1+x)}{1+x}dx=-\dfrac{1}{8}\zeta(3)$
