Question

Evaluate: $\int \frac{x{\tan }^{-1}x}{{(1+x^2)}^{3/2}}dx$

Answer 1

Put $ta{n}^{-1}x=t⟹\frac{1}{1+x^2}dx=dt$

$⟹\int \frac{xta{n}^{-1}x}{(1+x^2{)}^{3/2}}dx=\int \frac{ttant}{(1+ta{n}^{2}t{)}^{1/2}}dt$

$=\int \frac{ttant}{(se{c}^{2}t{)}^{1/2}}dt$

$=\int \frac{ttant}{sect}dt$

$=\int tsintdt$

$=t\int sintdt-\int (\frac{d\left(t\right)}{dt}\int sintdtdt$

$=-tcost-\int (-cost)dt+c$

$=-tcost+sint+c$

Now, as $tant=x$ from our substitution, $sint=\frac{x}{\sqrt{1+x^2}}$ and $cost=\frac{1}{\sqrt{1+x^2}}$

$⟹⟹\int \frac{xta{n}^{-1}x}{(1+x^2{)}^{3/2}}dx=-ta{n}^{-1}x\frac{1}{\sqrt{1+x^2}}+\frac{x}{\sqrt{1+x^2}}+c$

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