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
\(∫\frac{5x+3}{\sqrt{x^2+4x+10}} dx = \frac{5}{2}[2\sqrt{x^2+4x+10}]-7log|(x+2)+\sqrt{x^2+4x+10}|+C\)
\(=5\sqrt{x^2+4x+10} - 7 log|(x+2)+\sqrt{x^2+4x+10}|+C\)