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
\(∫\frac{x+3}{x^2-2x-5} dx = \frac{1}{2}log|x^2-2x-5|+\frac{4}{2\sqrt{6}} log|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}|+C\)
\(=\frac{1}{2}log|x^2-2x-5|+\frac{2}{\sqrt{6}} log|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}|+C\)