Question

$\int {\sin }^{-1}\left(\frac{2x}{1+x^2}\right)dx=f\left(x\right)-\log (1+x^2)+c$ then $f\left(x\right)=$

• A. $2x{\tan }^{-1}x$
• B. $-2x{\tan }^{-1}x$
• C. $x{\tan }^{-1}x$
• D. $-x{\tan }^{-1}x$

Answer 1

The correct option is A $2x{\tan }^{-1}x$

$\int si{n}^{-1}\left(\frac{2x}{1+x^2}\right)dx=f\left(x\right)-log(1+x^2)+c$

$\int si{n}^{-1}\left(\frac{2x}{1+x^2}\right)dx;letx=tan\theta$

$\int si{n}^{-1}\left(\frac{2tan\theta }{1+ta{n}^2\theta }\right)se{c}^2\theta d\theta$

$=2\int \theta se{c}^2\theta do$

$=2[\theta \int se{c}^2\theta d\theta -\int 1(\int se{c}^2\theta d\theta )d\theta ]$

$=2[\theta tan\theta -\int tan\theta d\theta +c]$

$2\theta tan\theta +2log\left|cos\theta \right|+c$

converting in to 'x'

$2xta{n}^{-1}x+2\left(log\frac{1}{\sqrt{1+x^2}}\right)+c$

$f\left(x\right)=2xta{n}^{-1}x$