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.

Hint: integrate by parts.
Let $u = \ln(x)\implies u' = \dfrac1x$ and $v' = \dfrac1{1 + x}\implies v = \ln(1 + x)$. Therefore,
$$\int\dfrac{\ln(x)}{1 + x}\,\mathrm dx = \ln(x)\ln(1 + x) - \int\dfrac{\ln(1 + x)}x\,\mathrm dx$$
