Question

$\int \frac{x+2}{(x^2+3x+3)\sqrt{x+1}}dx$ is equal to

• A. $\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x}{\sqrt{3(x+1)}}\right)$
• B. $\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{x}{\sqrt{3(x+1)}}\right)$
• C. $\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{x}{\sqrt{x+1}}\right)$
• D. none of these

Answer 1

The correct option is B $\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{x}{\sqrt{3(x+1)}}\right)$

Let $I=\int \frac{x+2}{(x^2+3x+3)\sqrt{x+1}}dx$
Put $x+1=t^2$ $\Rightarrow dx=2tdt$
$I=2\int \frac{t^2+1}{t^4+t^2+1}dt$
$=2\int \frac{1+(1/t{)}^2}{{(t-\frac{1}{t})}^2+3}dt$
Put $t-\frac{1}{t}=u$ $\Rightarrow 1+\frac{1}{t^2}=\frac{du}{dt}$
$=2\int \frac{du}{u^2+3}$
$=\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{u}{\sqrt{3}}\right)+C$
$=\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{3}}\right)+C$

$=\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{x}{\sqrt{3(x+1)}}\right)+C$

Hence, option B.