Question

If ${\sin }^{-1}\frac{2x}{1+x^2};{\cos }^{-1}\frac{1-x^2}{1+x^2};{\tan }^{-1}\frac{2x}{1-x^2}.$each is equal to $2{\tan }^{-1}x.$ ,then show that $\int 2{\tan }^{-1}x=2[x{\tan }^{-1}x-\frac{1}{2}\log (1+x^2)]$

Answer 1

Let $I=2\int {\tan }^{-1}xdx$

Applying integration by parts

$=2x{\tan }^{-1}x-2\int \frac{x}{x^2+1}dx$

Put $x^2+1=t$

$\Rightarrow 2xdx=dt$
$\Rightarrow I=2x{\tan }^{-1}x-\int \frac{dt}{t}$

$=2x{\tan }^{-1}x-\log t+C$

$I=2x{\tan }^{-1}x-\log (x^2+1)+C$