<s>[INST] Two definite integrals These integrals are closely related since $\frac{\pi^2}{8}=\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}$ and $G=\sum_{n\mathop=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^2}$. I'm not able to prove them though.

Show that
  $$\int_0^1 \frac{1}{\sqrt{x^2-1}}\log\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\right)\,\mathrm{d}x =\frac{\pi^2}{8}$$
  $$\int_1^{\infty} \frac{1}{x\sqrt{x^2-1}}\log\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\right)\,\mathrm{d}x =2G$$
where $G$ is Catalan's constant.

 [/INST] Note that $\cosh^{-1}{x}=\log\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\right)$ and $\frac{d}{dx}\cosh^{-1}{x}=\frac{1}{\sqrt{x^2-1}}$. Substituting $u=\cosh^{-1}{x}$ into the first integral, we get:
$$\int_0^1 \frac{1}{\sqrt{x^2-1}}\log\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\right)\,\mathrm{d}x=\int_{\frac{i\pi}{2}}^{0}udu=\frac12u^2\big |_{\frac{i\pi}{2}}^0=-\frac12(\frac{i\pi}{2})^2=\frac{\pi^2}{8}.$$
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