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**chisigma** With a little of patience You can arrive to the general expression...

$\displaystyle \displaystyle \int_{0}^{1} x^{n}\ \ln x\ dx = - \frac{1}{(n+1)^{2}}$ (1)

... and from (1) and the identity...

$\displaystyle \displaystyle \frac{1}{x^{2}-1} = -1 -x^{2} - x^{4} - ... - x^{2n} - ...$ (2)

... You have finally...

$\displaystyle \displaystyle \int_{0}^{1} \frac{\ln x}{x^{2} -1}\ dx = \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} = \frac{\pi^{2}}{8}$ (3)

Following a similar procedure You can solve also the second integral...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$