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

Equating the coefficients of $x^2$ and $x$ and constant term, we obtain

$B=-\frac{1}{10},A=\frac{5}{2}$, and $C=-\frac{24}{5}$
$\therefore \frac{2x-3}{(x+1)(x-1)(2x+3)}=\frac{5}{2(x+1)}-\frac{1}{10(x-1)}-\frac{24}{5(2x+3)}$
$\Rightarrow \int \frac{2x-3}{(x^2-1)(2x+3)}dx=\frac{5}{2}\int \frac{1}{(x+1)}dx-\frac{1}{10}\int \frac{1}{x-1}dx-\frac{24}{5}\int \frac{1}{(2x+3)}dx$
$=\frac{5}{2}\log |x+1|-\frac{1}{10}\log |x-1|-\frac{24}{5\times 2}\log |2x+3|$
$=\frac{5}{2}\log |x+1|-\frac{1}{10}\log |x-1|-\frac{12}{5}\log |2x+3|+C$