Can anyone help me in changing the integral into the given form: $\lim_{n \to \infty}n^{2} \Biggl(\ \ \int\limits_{0}^{1} \sqrt[n]{1+x^{n}} \ \text{dx}-1 \Biggr) = \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{2}}$
Once this is done, we know that the integral converges to $\frac{\pi^2}{12}$.
Added: One can generally see that $(1+x)^{a} = \sum\limits_{k=0}^{\infty} { a \choose k} x^{k}; \qquad x \in [0,1], \ a \in (0,1)$