Question

If $\int \frac{dx}{(x+100)\sqrt{x+99}}=f\left(x\right)+C,$ then $f\left(x\right)$ is equal to (where $C$ is integration constant)

• A. ${\tan }^{-1}\left(\sqrt{x+99}\right)$
• B. ${\tan }^{-1}\left(\sqrt[3]{3x+99}\right)$
• C. $2{\tan }^{-1}\left(\sqrt{x+99}\right)$
• D. $2{\sin }^{-1}\left(\sqrt{x+99}\right)$

Answer 1

The correct option is C $2{\tan }^{-1}\left(\sqrt{x+99}\right)$

Let, $I=\int \frac{dx}{(x+100)\sqrt{x+99}}$
Put $x+99=t^2\Rightarrow dx=2tdt$
$\Rightarrow I=\int \frac{2tdt}{(t^2+1)t}$
$=2\int \frac{1}{t^2+1}dt$
$=2{\tan }^{-1}t+C$
$=2{\tan }^{-1}\left(\sqrt{x+99}\right)+C$
$\therefore f\left(x\right)=2{\tan }^{-1}\left(\sqrt{x+99}\right)$