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 the above

Answer 1

Answer: (B)

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

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

Put $x+1=t^2\Rightarrow dx=2tdt$

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

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

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

$=2\int \frac{du}{u^2+{\left(\sqrt{3}\right)}^2}$

$(where,u=t-\frac{1}{t}\Rightarrow du=1+\frac{1}{t^2}dt)$

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

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

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