Question

If $\int \frac{1}{(1+x)\sqrt{x}}dx=f\left(x\right)+A$, where A is any arbitary constant, then the function f(x) is

• A. $2{\tan }^{-1}x$
• B. $2{\tan }^{-1}\sqrt{x}$
• C. $2{\cot }^{-1}\sqrt{x}$
• D. $lo{g}_e(1+x)$

Answer 1

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

Given,

$\int \frac{1}{(1+x)\sqrt{x}}dx$

put $u=\sqrt{x}$

$=\int \frac{2}{1+u^2}du$

$=2{\tan }^{-1}\left(u\right)$

$=2{\tan }^{-1}\left(\sqrt{x}\right)+C$

$=f\left(x\right)+A$

$\therefore f\left(x\right)=2{\tan }^{-1}\left(\sqrt{x}\right)$