Question

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

Answer 1

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

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

$\Rightarrow$ (2x-3) = A(x-1)(2x+3)+B(x+1)(2x+3)+C(x+1)(x-1)

$\Rightarrow$ (2x-3) = A(2x2+x-3)+B(2x2+5x+3)+C(x2-1)

$\Rightarrow$ (2x-3) = A(2A+2B+C)x2+(A+5B)x+(-3A+3B-C)

Equating the coefficients of x2 and x, we obtain

B = -$\frac{1}{10}$, A = $\frac{5}{2}$, and C = -$\frac{24}{5}$

∴ $\frac{2x-3}{(x+1)(x-1)(2x+3)}= \frac{5}{2(x+1)}\frac{1}{10(x-1)}+\frac{24}{5(2x+3)}$

$\Rightarrow\int\frac{2x-3}{(x+1)(x-1)(2x+3)}dx= \frac{5}{2}\int\frac{1}{(x+1)}dx-\frac{1}{10}\int\frac{1}{x-1}dx-\frac{24}{5}\int\frac{1}{(2x+3)}dx$

= $\frac{5}{2}\log\mid x+1\mid-\frac{1}{10}\log \mid x-1\mid-\frac{24}{5*2}\log\mid 2x+3\mid$

=$\frac{5}{2}\log\mid x+1\mid-\frac{1}{10}\log\mid x-1\mid-\frac{12}{5}\log\mid 2x+3 \mid+C$