Question

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

Answer 1

The correct answer is: $5\sqrt{x^2+4x+10} - 7 log|(x+2)+\sqrt{x^2+4x+10}|+C$
Let $5x+3 = A \frac{d}{dx}(x^2+4x+10)+B$
$⇒ 5x+3 = A(2x+4)+B$
Equating the coefficients of $x$ and constant term, we obtain
$2A = 5 ⇒ A = \frac{5}{2}$
$4A+B = 3 ⇒ B =-7$
$∴ 5x+3 = \frac{5}{2}(2x+4)-7$
$⇒ ∫\frac{5x+3}{\sqrt{x^2+4x+10}} dx = ∫\frac{\frac{5}{2}(2x+4)-7}{\sqrt{x^2+4x+10}} dx$
$=\frac{5}{2} ∫\frac{2x+4}{\sqrt{x^2+4x+10}} dx-7 ∫\frac{1}{\sqrt{x^2+4x+10}} dx$
Let $I_1 = ∫\frac{2x+4}{\sqrt{x^2+4x+10}} dx\,\, and\,\, I_2 = ∫\frac{1}{\sqrt{x^2+4x+10}} dx$
$∴ ∫\frac{5x+3}{\sqrt{x^2+4x+10}} dx = \frac{5}{2I_1}-7I_2 ...(1)$
Then, $I_1 = ∫\frac{2x+4}{\sqrt{x^2+4x+10}} dx$
Let $x^2+4x+10 = t$
$∴(2x+4)dx = dt$
$⇒ I_1 = ∫\frac{dt}{t} = 2\sqrt{t} = 2\sqrt{x^2+4x+10} ...(2)$
$I_2 = ∫\frac{1}{\sqrt{x^2+4x+10}} dx$
$= ∫\frac{1}{\sqrt{(x^2+4x+4)+6}} dx$
$= ∫\frac{1}{(x+2)^2+(\sqrt{6})^2} dx$
$= log |(x+2)\sqrt{x^2+4x+10}|$ ...(3)
Using equations (2) and (3) in (1), we obtain
$∫\frac{5x+3}{\sqrt{x^2+4x+10}} dx = \frac{5}{2}[2\sqrt{x^2+4x+10}]-7log|(x+2)+\sqrt{x^2+4x+10}|+C$
$=5\sqrt{x^2+4x+10} - 7 log|(x+2)+\sqrt{x^2+4x+10}|+C$