<s>[INST] Is this proof of the following integrals fine $\int_{0}^{1} \frac{\ln(1+x)}{x} dx$? $$\int_{0}^{1} \frac{\ln(1+x)}{x} dx = \int_{0}^{1} \frac1x\cdot (x-x^2/2 + x^3/3 -x^4/4 \ldots)dx = \int_{0}^{1}(1 - x/2 + x^2/3 - x^4/4 \dots)dx = 1-\frac1{2^2} + \frac1{3^3} - \frac{1}{4^4} \dots = \frac{\pi^2}{12}$$
Also, 
$$\int_{0}^{1} \frac{\ln(1-x)}{x}dx = -\int_{0}^{1}\frac1x \cdot (x+x^2/2 +x^3/3 \ldots)dx =- \int_{0}^{1}(1+x/2+x^2/3 \dots )dx = -( 1 + \frac1{2^2} + \frac1{3^3} \dots) = -\frac{\pi^2}{6}$$
Using these integral, I evaluate $$ \int_{0}^{1} \frac{\ln(x)}{1-x} $$
by letting $1-x=t$ 
$$ \int_{0}^{1} \frac{\ln(x)}{1-x} =\int_{0}^{1} \frac{\ln(1-t)}{t} dx = -\frac{\pi^2}{6}$$
However, I'm facing problem while integrating 
$$\int_{0}^{1} \frac{\ln(x)}{1+x}$$ using the same above way.
 [/INST] Let,
$\begin{align}I&=\int_0^1 \frac{\ln x}{1-x}\,dx\\
J&=\int_0^1 \frac{\ln x}{1+x}\,dx\\
\end{align}$
$\begin{align}I-J&=\int_0^1 \frac{2x\ln x}{1-x^2}\,dx\end{align}$
Perform the change of variable $y=x^2$,
$\begin{align}I-J&=\frac{1}{2}\int_0^1 \frac{\ln x}{1-x}\,dx\\
&=\frac{1}{2}I\end{align}$
Therefore, 
$J=\dfrac{1}{2}I$
If $I=-\dfrac{\pi^2}{6}$ then $J=-\dfrac{\pi^2}{12}$
</s>