I'm curious in knowing if there is any other method to solve this indefinite integral $$ \int \frac{\sqrt{x}}{1+x^{2}}\,dx $$
besides set $ \sqrt{x} = t $ and obtain $$ \int \frac{2t^{2}}{1+t^{4}}\,dt $$ where I reduce the problem to $$ \int \frac{t^{2}}{1+t^{4}}\,dt $$ and factorize $1+t^{4}=(t^{2}+\sqrt{2}t+1)(t^{2}-\sqrt{2}t+1)$ to reduce the integral to
$$ \int \frac{t^{2}}{(t^{2}+\sqrt{2}t+1)(t^{2}-\sqrt{2}t+1)}\,dt $$ This is not impossible with partial fraction but rather tedious.
Here is a shortcut
\begin{align} \int \frac{\sqrt{x}}{1+x^{2}}\,dx &= \int \frac{\frac1{\sqrt{x}}}{x+\frac1x}\,dx =\int \frac{d(\sqrt x- \frac1{\sqrt{x}}) + d(\sqrt x+\frac1{\sqrt{x}}) }{x+\frac1x}\\ &= \int \frac{d(\sqrt x- \frac1{\sqrt{x}}) }{(\sqrt x-\frac1{\sqrt x})^2+2} +\int \frac{d(\sqrt x+\frac1{\sqrt{x}}) }{(\sqrt x+\frac1{\sqrt x})^2-2}\\ &=\frac1{\sqrt2}\tan^{-1}\frac{ \sqrt x-\frac1{\sqrt x}}{\sqrt2} -\frac1{\sqrt2}\coth^{-1}\frac{ \sqrt x +\frac1{\sqrt x}}{\sqrt2}+C \end{align}