<s>[INST] How to prove $\int_0^1 \frac{\arctan( a x)}{x \sqrt{1 - x^2}} dx = \frac{\pi}{2} \ln (a + \sqrt{1 + a^2})$? I found the integral
$$\int_0^1 \frac{\arctan( a x)}{x \sqrt{1 - x^2}} dx = \frac{\pi}{2} \ln (a + \sqrt{1 + a^2})$$
in Gradshteyn and Ryzhik book and I try to solve it, but can't.
Taking the derivative, you get
$$\int_0^1 \frac{1}{(1 + a^2 x^2) \sqrt{1 - x^2}} dx = \frac{\pi}{2} \frac{1}{\sqrt{1 + a^2}}$$,
but this integral doesn't seem any easier. Have hints? Highschool methods would be best for me.
 [/INST] $$\large I(\alpha)=\int_0^1 \frac{\text{arctan }(\alpha x)}{x\sqrt{1-x^2}}$$
$$\large I'(\alpha)=\frac{\pi}{2}\frac{1}{\sqrt{1+\alpha^2}}$$
as per your question
$$ I(\alpha)=\frac{\pi}{2}\int\frac{1}{\sqrt{1+\alpha^2}}$$
$$ I(\alpha)=\frac{\pi}{2}\int\frac{\cosh\theta}{\sqrt{1+\sinh\theta^2}}$$
$$ I(\alpha)=\frac{\pi}{2}\int\frac{1}{1}$$
$$ I(\alpha)=\frac{\pi}{2}\theta+C$$
$$ I(\alpha)=\frac{\pi}{2}\text{arcsinh } \alpha+C$$
Now we find $C$.
$$ I(0)=\int_0^1 \frac{0}{x\sqrt{1-x^2}}=0=\frac{\pi}2 \times 0 +C$$
Therefore $C=0,$ and therefore 
$$ I(\alpha)=\frac\pi 2 \text{arcsinh } \alpha$$
$$ \large I(\alpha)=\frac\pi 2 (\log[\alpha+\sqrt{\alpha^2+1}])$$
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