Calculate using integration by parts $$\int \dfrac{x^2}{(x^2+1)^2}$$
I'm looking through some working for this question and it gives
$u=x/2, u'=1/2$
$v'=\dfrac{2x}{(x^2+1)^2}, v=\dfrac{-1}{x^2+1}$
I'm confused as to why these values for u and v are used and would appreciate it if someone could explain why
Notice, $$\int \frac{x^2}{(x^2+1)^2}\ dx$$$$=\frac{1}{2}\int \underbrace{x}_{I}\cdot \underbrace{\frac{2x}{(x^2+1)^2}}_{II}\ dx$$ Now, using integration by parts, $$=\frac{1}{2}\left(x\int \frac{2x}{(x^2+1)^2}\ dx-\int \left(\frac{d}{dx}(x)\cdot \int \frac{2x}{(x^2+1)^2}\ dx\right) dx\right)$$ $$=\frac{1}{2}\left(x\int \frac{d(x^2+1)}{(x^2+1)^2}-\int \left(\frac{d}{dx}(x)\cdot \int \frac{d(x^2+1)}{(x^2+1)^2}\right)\ dx\right)$$ $$=\frac{1}{2}\left(x\left(-\frac{1}{x^2+1}\right)-\int 1\cdot \left(-\frac{1}{x^2+1}\right)\ dx\right)$$
$$=\frac{1}{2}\left(-\frac{x}{x^2+1}\right)-\frac{1}{2}\int \left(-\frac{1}{x^2+1} \right) dx$$ $$=-\frac{x}{2(x^2+1)}+\frac{1}{2}\int \frac{1}{x^2+1}\ dx$$ $$=-\frac{x}{2(x^2+1)}+\frac{1}{2}\tan^{-1}(x)+C$$