integrate $\frac{\sqrt{\arctan\frac{1}{x}}}{1+x^2}$ I was trying to solve $\int \frac{\sqrt{\arctan\frac{1}{x}}}{1+x^2} dx$. I think I have the first step right, which is $\int \sqrt{\arctan\frac{1}{x}} d(\arctan x) $, but I'm not sure how to proceed. I considered integration by parts, but it seems to make the new integrand too complicated. please help. Thanks.

Sunstitute $u/\tan^{-1}(1/x)$.  Thus $x=\cot(u), dx=-\csc^2(u) du$.  You should find the trigonometric functions cancel out and the integral is then easy.
