Question

The value of $\int \frac{d^2}{d{x}^2}\left(ta{n}^{-1}x\right)dx$ is equal to

• A. $\frac{1}{1+x^2}$ + C
• B. $ta{n}^{-1}x$ + C
• C. x tan - $\frac{1}{2}log(1+x^2)+C$
• D. $\frac{1+x^2}{2}$ + C

Answer 1

The correct option is A $\frac{1}{1+x^2}$ + C

$Q\int \frac{d^2}{d{x}^2}\left({\tan }^{-1}x\right)dx$

$=\int \frac{d}{dx}\frac{d}{dx}\left({\tan }^{-1}x\right)dx$

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

$\int \frac{(1+x^2)\frac{d}{dx}1-1\frac{d}{dx}(1+x^2)}{(1+x^2{)}^2}dx$

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

put $\Rightarrow 1+x^2=t\Rightarrow 2xdx=dt$

$=\int \frac{-1}{t^2}dt=\frac{-t^{-2+1}}{-2+1}+c=\frac{1}{t}+c$

$=\frac{1}{1+x^2}+c$