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\displaystyle [br]\begin{align*} I(\alpha) = \int_0^{\frac{\pi}{2}} \frac{\arctan (\alpha \tan x)}{\tan x} \, \mathrm{d}x & \Rightarrow I'(\alpha) = \int_0^{\frac{\pi}{2}} \frac{\mathrm{d}x}{(\alpha \tan x)^2 + 1} \\ & \Rightarrow I'(\alpha) = \frac{\pi}{2(\alpha +1)} \\ & \Rightarrow I(\alpha) = \frac{\pi}{2}\ln (\alpha + 1) + \mathcal{C} \end{align*}
\displaystyle [br]\begin{align*} I(\alpha) = \int_0^{\frac{\pi}{2}} \frac{\arctan (\alpha \tan x)}{\tan x} \, \mathrm{d}x & \Rightarrow I'(\alpha) = \int_0^{\frac{\pi}{2}} \frac{\mathrm{d}x}{(\alpha \tan x)^2 + 1} \\ & \Rightarrow I'(\alpha) = \frac{\pi}{2(\alpha +1)} \\ & \Rightarrow I(\alpha) = \frac{\pi}{2}\ln (\alpha + 1) + \mathcal{C} \end{align*}
\displaystyle [br]\begin{equation*} \int_0^1 \frac{x^2 - 1}{\ln x} \, \mathrm{d}x \end{equation*}
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