Question

$\int \frac{dx}{(x+100)\sqrt{x+99}}=f\left(x\right)+c⟹f\left(x\right)=$

• A. $2(x+100{)}^{1/2}$
• B. $3(x+100{)}^{1/2}$
• C. $2{\tan }^{-1}\left(\sqrt{x+99}\right)$
• D. $2{\tan }^{-1}\left(\sqrt{x+100}\right)$

Answer 1

Answer: (C)

$2{\tan }^{-1}\left(\sqrt{x+99}\right)$

$\int \frac{dx}{(x+100)\sqrt{x+99}}$

$=2\int \frac{\frac{dx}{2\sqrt{x+99}}}{(1+(\sqrt{x+99}{)}^2)}$

$=2\int \frac{d\left(\sqrt{x+99}\right)}{(1+(\sqrt{x+99}{)}^2)}$

$=2{\tan }^{-1}\left(\sqrt{x+99}\right)+c$ [ Where $c$ is integrating constant]

Comparing this with gives equation we get,

$f\left(x\right)=2{\tan }^{-1}\left(\sqrt{x+99}\right)$.