Question

Integrate the function: $\frac{x+3}{x^2-2x-5}$

Answer 1

The correct answer is: $=\frac{1}{2}log|x^2-2x-5|+\frac{2}{\sqrt{6}} log|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}|+C$
Let $(x+3) = A\frac{d}{dx}(x^2-2x-5)+B$(x+3) = A(2x-2)+B
Equating the coefficients of $x$ and constant term on both sides, we obtain
$2A = 1 ⇒ A = \frac{1}{2}$
$-2A+B = 3 ⇒ B=4$
$∴ (x+3) = \frac{1}{2}(2x-2)+4$
$⇒ ∫\frac{x+3}{x^2-2x-5} dx = ∫\frac{\frac{1}{2}(2x-2)+4}{x^2-2x-5} dx$
$=\frac{1}{2} ∫\frac{2x-2}{x^2-2x-5} dx + 4 ∫\frac{1}{x2-2x-5} dx$
Let $I_1 = ∫\frac{2x-2}{x^2-2x-5} dx\, and\, I_2 = ∫\frac{1}{x^2-2x-5} dx$
$∴ ∫\frac{x+3}{(x^2-2x-5)}dx = \frac{1}{2}I_1+4I_2 ...(1)$
Then, $I_1 = ∫\frac{2x-2}{x^2-2x-5} dx$
Let $x^2-2x-5 = t$
$⇒(2x-2)dx = dt$
$⇒ I_1 = ∫\frac{dt}{t} = log|t|=log|x^2-2x-5| ...(2)$
$I_2 = ∫\frac{1}{x^2-2x-5} dx$
$= ∫\frac{1}{(x^2-2x+1)-6} dx$
$= ∫\frac{1}{(x-1)^2+(\sqrt{6})^2} dx$
$= \frac{1}{2\sqrt{6}} log(\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}})$ ...(3)
Substituting (2) and (3) in (1), we obtain
$∫\frac{x+3}{x^2-2x-5} dx = \frac{1}{2}log|x^2-2x-5|+\frac{4}{2\sqrt{6}} log|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}|+C$
$=\frac{1}{2}log|x^2-2x-5|+\frac{2}{\sqrt{6}} log|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}|+C$