Question

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

• A. $\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{x}{x+1}\right)+C$
• B. $\frac{2}{\sqrt{3}}{\tan }^{-1}\left[\frac{x}{\sqrt{3\left(x+1\right)}}\right]+C$
• C. $\frac{2}{\sqrt{3}}{\tan }^{-1}\left[\frac{x}{{\left(x+1\right)}^2}\right]+C$
• D. None of the above

Answer 1

Answer: (B)

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