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.

$$
u = \arctan\left(\frac{1}{x}\right) \rightarrow du = \frac{1}{1+\frac{1}{x^2}} \cdot \frac{-1}{x^2} dx = \frac{-1}{1+x^2} dx
$$
$$
\int \frac{\sqrt{\arctan\frac{1}{x}}}{1+x^2} dx = \int -\sqrt{u} du = -\frac{2}{3}u^{3/2} + C = -\frac{2}{3}\left(\arctan\left(\frac{1}{x}\right)\right)^{3/2} + C
$$
So:
$$
\int \frac{\sqrt{\arctan\frac{1}{x}}}{1+x^2} dx = -\frac{2}{3}\left(\arctan\left(\frac{1}{x}\right)\right)^{3/2} + C
$$
