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A logarithmic integral

Let \alpha \in \mathbb{R}. Prove that:

    \[\int_{0}^{\infty} \frac{\arctan \alpha \sin^2 x}{x^2}\, \mathrm{d}x = \frac{\pi}{\sqrt{2}} \cdot \frac{\alpha}{\sqrt{1+ \sqrt{1+\alpha^2}}}\]

Solution

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