Question

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

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

Answer 1

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

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

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

Therefore, $\int \frac{2(t^2+1)tdt}{\left[\right(t^2-1{)}^2+3t^2].t}=\int \frac{2(t^2+1)dt}{\left[\right(t^4+t^2+1)}=\int \frac{2(t^2+1)dt}{\left[\right(t^2-t+1\left)\right(t^2+t+1)}$

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

$=\frac{2}{\sqrt{3}}[{\tan }^{-1}\left(\frac{2}{\sqrt{3}}(t+\frac{1}{2})\right)+{\tan }^{-1}\left(\frac{2}{\sqrt{3}}(t-\frac{1}{2})\right)]=\frac{2}{\sqrt{3}}{\tan }^{-1}\left[\frac{\frac{4t}{\sqrt{3}}}{1-\frac{4}{3}(t^2-\frac{1}{4})}\right]$

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

Therefore, $I=\frac{2}{\sqrt{3}}{\cot }^{-1}\left(\frac{\sqrt{3(x+1)}}{x}\right)+c$