Indefinite integral: $\int \frac{x\mathrm{ln}x}{(1+x^2)^2}\mathrm{d}x$ I am stuck with the following integral: $$\int \frac{x\mathrm{ln}x}{(1+x^2)^2}\mathrm{d}x$$
I'm sure integration by parts has to be involved here, but I cannot find the proper function to use the formula. Any ideas would be helpful.

Try integration by parts with $u = \ln(x)$, $dv = \frac{x}{(1+x^2)^2}$. You can easily get $du$ and you can solve for $v$ via a pretty straightforward antidifferentiation of $dv.$ You'll get $$\int \frac{x\mathrm{ln}x}{(1+x^2)^2}\mathrm{d}x = -\frac{\ln(x)}{2(1+x^2)}+\frac{1}{2} \int \frac{1}{x(1+x^2)}$$ Now you can use partial fraction decomposition on $\frac{1}{x(1+x^2)}$ to rewrite $\frac{1}{x(1+x^2)} = \frac{A}{x}+\frac{Bx+C}{1+x^2}$ for some constants $A,B,C$. The rest of the integral should be easy.
