Question

Integrate the function: $\frac{5x+3}{\sqrt{x^2+4x+10}}$

Answer 1

$\int \frac{5x+3}{\sqrt{x^2+4x+10}}dx$
$=5\int \frac{(x+\frac{3}{5})}{\sqrt{x^2+4x+10}}dx$
$=\frac{5}{2}\int \frac{2x+\frac{6}{5}}{\sqrt{x^2+4x+10}}dx$
$=\frac{5}{2}\int \frac{2x+4-4+\frac{6}{5}}{\sqrt{x^2+4x+10}}dx$
$=\frac{5}{2}\int \frac{2x+4-\frac{14}{5}}{\sqrt{x^2+4x+10}}dx$
$=\frac{5}{2}\int \frac{2x+4}{\sqrt{x^2+4x+10}}dx+\frac{5}{2}\int \frac{-\frac{14}{5}}{\sqrt{x^2+4x+10}}dx$
$\stackrel{\underset{}{=\frac{5}{2}\int \frac{2x+4}{\sqrt{x^2+4x+10}}dx}}{I_1}$
$\stackrel{\underset{}{-7\int \frac{dx}{\sqrt{x^2+4x+10}}dx}}{I_2}$
For $I_1$
$I_1=\frac{5}{2}\int \frac{2x+4}{\sqrt{x^2+4x+10}}dx$
Let $x^2+4x+10=t^2$
Differentiate both sides w.r.t.x
$2x+4=2t\frac{dt}{dx}\Rightarrow (2x+4)dx=2tdt$
$\Rightarrow I_1=\frac{5}{2}\int \frac{2x+4}{\sqrt{x^2+4x+10}}dx$
$\Rightarrow I_1=\frac{5}{2}\int \frac{2tdt}{t}$
$\Rightarrow I_1=5\int 1.dt$
$\Rightarrow I_1=5t+C_1$
$\Rightarrow I_1=5\sqrt{x^2+4x+10}+C_1$
For $I_2$
$I_2=\int \frac{7}{\sqrt{x^2+4x+10}}.dx$
$\Rightarrow I_2$
$=7\int \frac{1}{\sqrt{x^2+4x+2^2-2^2+10}}.dx$
$\Rightarrow I_2=7\int \frac{1}{\sqrt{(x+2{)}^2+6}}.dx$
$\Rightarrow I_2=7\frac{1}{\sqrt{(x+2{)}^2+6}}.dx$
$\Rightarrow I_2=7\int \frac{1}{\sqrt{(x+2{)}^2+(\sqrt{6}{)}^2}}.dx$
$\Rightarrow I_2=7\left[\log |x+2+\sqrt{(x+2{)}^2+(\sqrt{6})}|\right]+C_2$
$[\because \int \frac{dx}{\sqrt{x^2+a^2}}=\log |x+\sqrt{x^2+a^2}|+C]$
$\Rightarrow I_2=7\log |x+2+\sqrt{x^2+4x+4+6}|+C_2$
$\Rightarrow I_2=7\log |x+2+\sqrt{x^2+4x+10}|+C_2$
$\int \frac{5x+3}{\sqrt{x^2+4x+10}}dx$
$=I_1-I_2$
$=5\sqrt{x^2+4x+10}+C_1$
$-7\log |x+2+\sqrt{x^2+4x+10}|-C_2$
$=5\sqrt{x^2+4x+10}-7\log |x+2+\sqrt{x^2+4x+10}+C$
Where $C=C_1-C_2$