Question

Evaluate: $\int \frac{1}{(x+2)\sqrt{x+1}}dx$

• A. $2{\tan }^{-1}\left(\sqrt{x+1}\right)+c$
• B. ${\tan }^{-1}\left(\sqrt{x-1}\right)+c$
• C. $2{\tanh }^{-1}\left(\sqrt{x+1}\right)-c$
• D. ${\tan }^{-1}\left(\sqrt{x+1}\right)+c$

Answer 1

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

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

$du=\frac{1}{2\sqrt{x+1}}dx$

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

Hence the correct option is $A,2{\tan }^{-1}\left(\sqrt{x+1}\right)+C$