Question

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

Answer 1

Let $5x+3=A\frac{d}{dx}(x^2+4x+10)+B$
$\Rightarrow 5x+3=A(2x+4)+B$
Equating the coefficients of $x$ and constant term, we obtain
$2A=5\Rightarrow A=\frac{5}{2}$
& $4A+B=3\Rightarrow B=-7$
$\therefore 5x+3=\frac{5}{2}(2x+4)-7$
$\Rightarrow \int \frac{5x+3}{\sqrt{x^2+4x+10}}dx=\int \frac{\frac{5}{2}(2x+4)-7}{\sqrt{x^2+4x+10}}dx$
$=\frac{5}{2}\int \frac{2x+4}{\sqrt{x^2+4x+10}}dx-7\int \frac{1}{\sqrt{x^2+4x+10}}dx$
Let $I_1=\int \frac{2x+4}{\sqrt{x^2+4x+10}}dx$ and $I_2=\int \frac{1}{\sqrt{x^2+4x+10}}dx$
$\therefore \int \frac{5x+3}{\sqrt{x^2+4x+10}}dx=\frac{5}{2}{I}_1-7I_2$ .......... (1)
Then, $I_1=\int \frac{2x+4}{\sqrt{x^2+4x+10}}dx$
Let $x^2+4x+10=t$
$\therefore (2x+4)dx=dt$
$\Rightarrow I_1=\int \frac{dt}{t}=2\sqrt{t}=2\sqrt{x^2+4x+10}$ ......(2)
$I_2=\int \frac{1}{\sqrt{x^2+4x+10}}dx$
$=\int \frac{1}{\sqrt{(x^2+4x+4)+6}}dx$
$=\int \frac{1}{(x+2{)}^2+(\sqrt{6}{)}^2}dx$
$=\log |(x+2)\sqrt{x^2+4x+10}|$ .............(3)
Using equation (2) and (3) in (1), we obtain
$\int \frac{5x+3}{\sqrt{x^2+4x+10}}dx=\frac{5}{2}\left[2\sqrt{x^2+4x+10}\right]-7\log |(x+2)+\sqrt{x^2+4x+10}|+C$
$=5\sqrt{x^2+4x+10}-7\log |(x+2)+\sqrt{x^2+4x+10}|+C$