Question

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

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

Answer 1

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

Let $I=\int \frac{x+2}{(x^2+3x+3)\sqrt{x+1}}dx$

Put $x+1=t^2$

$dx=2t.dt$

$\Rightarrow$ $I=2\int \frac{(x+1)+1}{\left[\right(x^2+2x+1)+x+2]\sqrt{x+1}}$

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

Put $x+1=t^2$

$dx=2t.dt$

$=2\int \frac{t^2+1}{t^4+t^2+1}dt$

$=2\int \frac{1+{\left(\frac{1}{t}\right)}^2}{{(t-\frac{1}{t})}^2+3}dt$

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

$=\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{\sqrt{x+1}-\frac{1}{\sqrt{x+1}}}{\sqrt{3}}\right)+c$ [ Since, $t^2=x+1$ then, $t=\sqrt{x+1}$ ]

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

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