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 = \(\frac{1}{2}\)

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

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

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

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