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