Question

Integrate the function $\frac{x+2}{\sqrt{x^2+2x+3}}$

Answer 1

$\int \frac{(x+2)}{\sqrt{x^2+2x+3}}dx=\frac{1}{2}\int \frac{2(x+2)}{\sqrt{x^2+2x+3}}dx$
$=\frac{1}{2}\int \frac{2x+4}{\sqrt{x^2+2x+3}}dx$
$=\frac{1}{2}\int \frac{2x+2}{\sqrt{x^2+2x+3}}dx+\frac{1}{2}\int \frac{2}{\sqrt{x^2+2x+3}}dx$
$=\frac{1}{2}\int \frac{2x+2}{\sqrt{x^2+2x+3}}dx+\int \frac{1}{\sqrt{x^2+2x+3}}dx$
Let $I_1=\int \frac{2x+2}{\sqrt{x^2+2x+3}}dx$ and $I_2=\int \frac{1}{\sqrt{x^2+2x+3}}dx$
$\therefore \int \frac{x+2}{\sqrt{x^2+2x+3}}dx=\frac{1}{2}{I}_1+I_2$ .........(1)
Then, $I_1=\int \frac{2x+2}{\sqrt{x^2+2x+3}}dx$
Let $x^2+2x+3=t$
$\Rightarrow (2x+2)dx=dt$
$I_1=\int \frac{dt}{\sqrt{t}}=2\sqrt{t}=2\sqrt{x^2+2x+3}$ ........ (2)
$I_1=\int \frac{1}{\sqrt{x^2+2x+3}}dx$
$\Rightarrow x^2+2x+3=x^2+2x+1+2=(x+1{)}^2+(\sqrt{2}{)}^2$
$\therefore I_2=\int \frac{1}{\sqrt{(x+1{)}^2+(\sqrt{2}{)}^2}}dx=\log |(x+1)+\sqrt{x^2+2x+3}|$ ......... (3)
Using equation (2) and (3) in (1), we obtain
$\int \frac{x+2}{\sqrt{x^2+2x+3}}dx=\frac{1}{2}\left[2\sqrt{x^2+2x+3}\right]+\log |(x+1)+\sqrt{x^2+2x+3}|+C$
$=\sqrt{x^2+2x+3}+\log |(x+1)+\sqrt{x^2+2x+3}|+C$