Question

Integrate the rational functions.
$\int \frac{2x-3}{(x^2-1)(2x+3)}dx.$

Answer 1

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

On comparing the coefficients of $x^2,x$ and constant term on both sides, we get

$2A+2B+C=0......\left(i\right)5A+B=2\Rightarrow B=2-5A....\left(ii\right)\text{and}3A-3B-C=-\mathrm{3......}\left(iii\right)\text{On putting the value of B in Eqs.}\left(i\right)\text{and}\left(iii\right),\text{we get}2A+2(2-5A)+C=0\Rightarrow 2A+4-10A+C=0\Rightarrow -8A+C=-4......\left(iv\right)\text{and}3A-3(2-5A)-C=-3\Rightarrow 3A-6+15A-C=-3\Rightarrow 18A-C=3......\left(v\right)\text{On adding Eqs.}\left(iv\right)\text{and}\left(v\right),\text{we get}10A=-1\Rightarrow A=\frac{-1}{10}\text{On putting the value of A in Eq.}\left(ii\right),\text{we get}B=2-5(-\frac{1}{10})\Rightarrow B=2+\frac{1}{2}=\frac{5}{2}\text{On putting the values of A and B in Eq.}\left(i\right),\text{we get}2(-\frac{1}{10})+2\left(\frac{5}{2}\right)+C=0\Rightarrow -\frac{1}{5}+5+C=0\Rightarrow C=\frac{24}{5}\therefore A=-\frac{1}{10},B=\frac{5}{2}\text{and}C=-\frac{24}{5}\therefore \int \frac{2x-3}{(x^2-1)(2x+3)}dx=\int \frac{(-1)}{10(x-1)}dx+\frac{5}{2}\int \frac{1}{x+1}dx-\frac{24}{5}\int \frac{1}{2x+3}dx=-\frac{1}{10}log|x-1|+\frac{5}{2}log|x+1|-\frac{24}{5}\frac{log|2x+3|}{2}+C=\frac{5}{2}log|x+1|-\frac{1}{10}log|x-1|-\frac{12}{5}log|2x+3|+C$