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.
 A: 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.
A: $$\int \dfrac{x\ln x}{(1+x^2)^2} \mathrm{dx}=\int\dfrac{x}{(1+x^2)^2}\ln x\mathrm{dx}=\dfrac{1}{2}\int\dfrac{2x}{(1+x^2)^2}\ln x\mathrm{dx}=\dfrac{1}{2}\int 2x(1+x^2)^{-2}\ln x\mathrm{dx}$$.
From here you can use partial integration with by taking the derivative of $\ln x$ and a primitive function of the other term which is of the form $f'f^n$. You will get
$$\int 2x(1+x^2)^{-2}\ln x\mathrm{dx}=\dfrac{1}{2}\left(-(1+x^2)^{-1}\ln x-\int -(1+x^2)^{-1}\dfrac{1}{x}\mathrm{dx}\right)=$$
$$
\dfrac{1}{2}\left(\dfrac{-\ln x}{1+x^2}-\int\dfrac{-1}{x+x^3} \mathrm{dx}\right).
$$
From here it is easy, write the denominator as $x(x^2+1)$ and use partial fractions.
