Ikunupe6v

2022-01-05

I am in the middle of a problem and having trouble integrating the following integral:
${\int }_{-1}^{1}\frac{1}{{\left(1+{x}^{2}\right)}^{2}}dx$

puhnut1m

Expert

Try making a substitution $x=\mathrm{tan}u$. Notice then that
${\left(1+{x}^{2}\right)}^{2}={\left(1+{\mathrm{tan}}^{2}u\right)}^{2}={\left({\mathrm{sec}}^{2}u\right)}^{2}={\left({\mathrm{sec}}^{2}u\right)}^{2}={\mathrm{sec}}^{4}u$
and
$dx={\mathrm{sec}}^{2}udu$
So the indefinite integral is now
$\int \frac{1}{{\mathrm{sec}}^{2}u}du=\int {\mathrm{cos}}^{2}udu$
This new integrand should be easier to integrate. Just remember to change your limits to get the proper evaluation.

Robert Pina

Expert

If we add and subtract ${x}^{2}$ in the numerator, we can integrate the first integral immediately
$\int \frac{1}{{\left(1+{x}^{2}\right)}^{2}}dx=\int \frac{1}{1+{x}^{2}}dx-\int \frac{{x}^{2}}{{\left(1+{x}^{2}\right)}^{2}}dx$
$\mathrm{arctan}x-\int x\frac{x}{{\left(1+{x}^{2}\right)}^{2}}dx$
and the second integral by parts:
$\int x\frac{x}{{\left(1+{x}^{2}\right)}^{2}}dx=x\left(-\frac{1}{2\left(1+{x}^{2}\right)}\right)+\int \frac{1}{2\left(1+{x}^{2}\right)}dx$
$=-\frac{x}{2\left(1+{x}^{2}\right\}+\frac{1}{2}\mathrm{arctan}x}$

user_27qwe

Expert

$\frac{1}{\left(1+{x}^{2}{\right)}^{2}}=\left(\frac{1}{1+{x}^{2}}-\frac{{x}^{2}}{1+{x}^{2}{\right)}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{1+{x}^{2}}+\frac{x}{2}\left(\frac{1}{1+{x}^{2}}{\right)}^{\prime }\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{1+{x}^{2}}+\left(\frac{1}{2}\frac{x}{1+{x}^{2}}\right)-\frac{1}{2}\frac{1}{1+{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(\mathrm{arctan}x+\frac{x}{1+{x}^{2}}{\right)}^{\prime }$