Question

Prove that
$\int \frac{\left(x{\tan }^{-1}x\right)}{(1+x^2{)}^{3/2}}dx$

Answer 1

Ref.Image.

$\int \frac{xta{n}^1\left(x\right)}{(1+x^2{)}^{3/2}}dx.$

Put $ta{n}^{-1}x=t$ $\therefore x=tant$

Differentiating $\therefore \frac{1}{1+x^2}dx.=dt$

$=\int \frac{tant.\left(t\right)}{\sqrt{1+x^2}}dt$ $(cost=\frac{1}{\sqrt{1+x^2}})$

$=\int \frac{tant.\left(t\right)}{\sqrt{1+ta{n}^{2}t}}dt$ $(\because 1+ta{n}^{2}t=se{c}^{2}t)$

$=\int \frac{t.tant}{sect}dt$ $(sint=\frac{x}{\sqrt{1+x^2}})$ (from figure)

$=\int t.sint.dt$

Applying Integration by parts

$=t.\int sint.dt-\int (\frac{d\left(t\right)}{dt}\int sint.dt)dt$

$=-t.cost-\int -cost.dt$

$=-t.cost+sint+c$

$=-ta{n}^{-1}x.\frac{1}{\sqrt{1+x^2}}+\frac{x}{\sqrt{1+x^2}}+c$

$=-\frac{ta{n}^{-1}x}{\sqrt{1+x^2}}+\frac{x}{\sqrt{1+x^2}}+c$