# Show that

Question:

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

Solution:

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

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

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

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

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

Equating the coefficients of $x^{2}$ and $x$. we obtain

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

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

$\Rightarrow \int \frac{2 x-3}{\left(x^{2}-1\right)(2 x+3)} d x=\frac{5}{2} \int \frac{1}{(x+1)} d x-\frac{1}{10} \int \frac{1}{x-1} d x-\frac{24}{5} \int \frac{1}{(2 x+3)} d x$

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

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