Question

Integrate the rational function: $\frac{3x+5}{x^3-x^2-x+1}$

Answer 1

$\frac{3x+5}{x^3-x^2-x+1}$ = $\frac{3x+5}{(x-1)2(x+1)}$

Let $\frac{3x+5}{(x-1)^2(x+1)}$ = $\frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+1)}$

3x+5 = A(x-1)(x+1)+B(x+1)+C(x-1)2

3x+5 = A(x2-1)+B(x+1)+C(x2+1-2x) ...(1)
Substituting x = 1 in equation (1), we obtain
B = 4
Equating the coefficients of x2 and x, we obtain
A + C = 0
B − 2C = 3
On solving, we obtain

A = -$\frac{1}{2}$ and C = $\frac{1}{2}$

∴ $\frac{3x+5}{(x-1)^2(x+1)} = \frac{-1}{2(x-1)}+\frac{4}{(x-1)^2}+\frac{1}{2(x+1)}$

$\Rightarrow \int \frac{3x+5}{(x-1)^2(x+1)}dx = -\frac{1}{2}\int\frac{1}{x-1}dx+4\int\frac{1}{(x-1)^2}dx+\frac{1}{2}\int\frac{1}{(x+1)}dx$

=$-\frac{1}{2}\log\mid x-1\mid+4\bigg(\frac{-1}{x-1}\bigg)+\frac{1}{2}\log\mid x+1\mid+C$

=$\frac{1}{2}\log\mid\frac{x+1}{x-1}\mid-\frac{4}{(x-1)}+C$