Question

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

Answer 1

$\int \frac{x+2}{\sqrt{x^2+2x+3}}dx$
$=\frac{1}{2}\int \frac{2x+4}{\sqrt{x^2+2x+3}}dx$
$=\frac{1}{2}\int \frac{2x+2+2}{\sqrt{x^2+2x+3}}dx$
$=\frac{1}{2}\int \frac{2x+2}{\sqrt{x^2+2x+3}}dx$
$+\frac{1}{2}\int \frac{2}{\sqrt{x^2+2x+3}}dx$
$=\stackrel{\underset{}{=\frac{1}{2}\int \frac{2x+2}{\sqrt{x^2+2x+3}dx}}}{i_1}$
$\stackrel{\underset{}{+\int \frac{dx}{\sqrt{x^2+2x+3}}dx}}{i_2}$
For $I_1$
$I_1=\frac{1}{2}\int \frac{2x+2}{\sqrt{x^2+2x+3}}.dx$
Let $x^2+2x+3=t^2$
Differential both sides w.r.t.x
$2x+2=25\frac{dt}{dx}\Rightarrow (2x+2)dx=2tdt$
$\Rightarrow I_1=\frac{1}{2}\int \frac{2tdt}{t}$
$\Rightarrow I_1=\int 1.dt$
$\Rightarrow I_1=t+C_1$
$\Rightarrow I_1=\sqrt{x^2+2x+3}+C_1$
For $I_2$
$I_2=\int \frac{1}{\sqrt{x^2+2x+3}}.dx$
$\Rightarrow I_2=\int \frac{1}{\sqrt{x^2+2x+1+2}}.dx$
$\Rightarrow I_2=\int \frac{1}{\sqrt{(x+1{)}^2+2}}.dx$
$\Rightarrow I_2=\int \frac{1}{\sqrt{(x+1{)}^2+(\sqrt{2}{)}^2}}.dx$
$\Rightarrow I_2=\log |x+1+\sqrt{(x+1{)}^2+(\sqrt{2}{)}^2}|+C_2$
$[\because \int \frac{dx}{\sqrt{x^2+a^2}}=\log |x+\sqrt{x^2+a^2}|+C]$
$\Rightarrow I_2=\log |x+1+\sqrt{x^2+2x+1+2}|+C_2$
$\Rightarrow I_2=\log |x+1+\sqrt{x^2+2x+3}|+C_2$
$\int \frac{(x+2)}{\sqrt{x^2+2x+3}}.dx$
$=I_1+I_2$
$=\sqrt{x^2+2x+3}+C_1+\log |x+1+\sqrt{x^2+2x+3}|+C_2$
$=\sqrt{x^2+2x+3}+\log |x+1+\sqrt{x^2+2x+3}|+C$
Where $C=C_1+C_2$