2022-01-05

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

nghodlokl

Expert

$\int \frac{dx}{{\left({x}^{2}+1\right)}^{2}}$
Set $x\phantom{\rule{0.222em}{0ex}}=\mathrm{tan}\left(u\right)$ and $dx={\mathrm{sec}}^{2}\left(u\right)du$. Then ${\left({x}^{2}+1\right)}^{2}={\left({\mathrm{tan}}^{2}\left(u\right)+1\right)}^{2}={\mathrm{sec}}^{4}\left(u\right)$ and $u=\mathrm{arctan}\left(x\right)$
$=\int {\mathrm{cos}}^{2}\left(u\right)du=\frac{1}{2}\int \mathrm{cos}\left(2u\right)du+\frac{1}{2}\int 1du$
$=\frac{u}{2}+\frac{1}{4}\mathrm{sin}\left(2u\right)+C$
$=\frac{{x}^{2}\mathrm{arctan}\left(x\right)+x+\mathrm{arctan}\left(x\right)}{2{x}^{2}+2}+C$

Laura Worden

Expert

There is a faster way. Substitute

thus

And remembering that
$\frac{1}{{\mathrm{sec}}^{2}}={\mathrm{cos}}^{2}$
your integral is simply
$\int {\mathrm{cos}}^{2}\left(z\right)dz$
Which is trivial and left to you.
The tangent/secant substitution is a great technique which most of people ignore. Study it, and you will solve lots of awesome integrals!
Final result:
$\frac{{x}^{2}\mathrm{arctan}\left(x\right)+x+\mathrm{arctan}\left(x\right)}{2{x}^{2}+2}$

star233

Expert

Let us find $\frac{d\left[\left(ax+b\right)/\left({x}^{2}+1{\right)}^{n}\right]}{dx}$
$=\frac{-a{x}^{2}-2bx+a}{\left({x}^{2}+1{\right)}^{n+1}}\phantom{\rule{0ex}{0ex}}=\frac{-a\left({x}^{2}+1\right)-2bx+2a}{\left({x}^{2}+1{\right)}^{n+1}}\phantom{\rule{0ex}{0ex}}=-\frac{a}{\left({x}^{2}+1{\right)}^{n}}+\frac{2a-2bx}{\left({x}^{2}+1{\right)}^{n+1}}$
Integrating both sides,
$b=0⇒2\int \frac{dx}{\left({x}^{2}+1{\right)}^{n+1}}=\int \frac{dx}{\left({x}^{2}+1{\right)}^{n}}+\frac{x}{\left({x}^{2}+1{\right)}^{n}}\phantom{\rule{0ex}{0ex}}n=1⇒2\int \frac{dx}{\left({x}^{2}+1{\right)}^{2}}=\int \frac{dx}{{x}^{2}+1}+\frac{x}{\left({x}^{2}+1\right)}$

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