Julia White

2022-01-03

How to calculate
$\int \frac{{x}^{2}}{{\left({x}^{2}+1\right)}^{2}}dx$

Terry Ray

First, you used parts: with $u=x$ and $dv=\frac{x}{{\left({x}^{2}+1\right)}^{2}}dx$, you got
$\int \frac{{x}^{2}}{{\left({x}^{2}+1\right)}^{2}}dx=xv-\int vdx$ (1)
From here, I suggested the substitution $t={x}^{2}$ in order to integrate dv and get v. Doing this, we see that $dt=2xdx$, so $xdx=\frac{1}{2}dt$. Thus
$v=\int dv=\int \frac{x}{{\left({x}^{2}+1\right)}^{2}}dx=\frac{1}{2}\int \frac{dt}{{\left(t+1\right)}^{2}}$
This latter integral is easy: if you can't guess it, let $s=t+1$. In any case, we get
$v=-\frac{1}{2\left(t+1\right)}=-\frac{1}{2\left({x}^{2}+1\right)}$
Going back to (1),
$\int \frac{{x}^{2}}{{\left({x}^{2}+1\right)}^{2}}dx=-\frac{x}{2\left({x}^{2}+1\right)}+\frac{1}{2}\int \frac{dx}{{x}^{2}+1}$
and you should recognize the final integral as $\mathrm{arctan}x$. Putting it all together, the answer is
$\frac{1}{2}\mathrm{arctan}x-\frac{x}{2\left({x}^{2}+1\right)}+C$

Daniel Cormack

HINT:
Use trigonometric substitution $x=\mathrm{tan}\theta$
$\int \frac{{x}^{2}}{{\left({x}^{2}+1\right)}^{2}}dx=\frac{1}{2}\int \left(1-\mathrm{cos}2\theta \right)d\theta$
$\mathrm{sin}2\theta =\frac{2\mathrm{tan}\theta }{1+{\mathrm{tan}}^{2}\theta }$

Vasquez

$\begin{array}{}\int \frac{{x}^{2}}{\left({x}^{2}+1{\right)}^{2}}dx\\ =x\int \frac{x}{\left({x}^{2}+1{\right)}^{2}}dx-\int \left[\frac{dx}{dx}\cdot \int \frac{x}{\left({x}^{2}+1{\right)}^{2}}dx\right]dx\\ =x\cdot \frac{-1}{1+{x}^{2}}+\int \frac{dx}{1+{x}^{2}}=...\end{array}$