Question

The value of $\int \frac{x+2}{(x^2+3x+3)\sqrt{x+1}}dx$ is (where $C$ is integration constant)

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

Answer 1

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

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