How to prove $\int_0^1 \frac{\arctan( a x)}{x \sqrt{1 - x^2}} dx = \frac{\pi}{2} \ln (a + \sqrt{1 + a^2})$?

Question

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.

Answer 1

$$\large I(\alpha)=\int_0^1 \frac{\text{arctan }(\alpha x)}{x\sqrt{1-x^2}}$$ $$\large I'(\alpha)=\frac{\pi}{2}\frac{1}{\sqrt{1+\alpha^2}}$$ 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 $$ I(\alpha)=\frac\pi 2 \text{arcsinh } \alpha$$ $$ \large I(\alpha)=\frac\pi 2 (\log[\alpha+\sqrt{\alpha^2+1}])$$