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Integrate the rational functions
$\frac{2 x-3}{\left(x^2-1\right)(2 x+3)}$
$\frac{2 x-3}{\left(x^2-1\right)(2 x+3)}$
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Let $\frac{2 x-3}{\left(x^2-1\right)(2 x+3)}=\frac{2 x-3}{(x-1)(x+1)(2 x+3)}$
$\equiv \frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{2 x+3}$
$\Rightarrow 2 x-3=\mathrm{A}(x+1)(2 x+3)+\mathrm{B}(x-1)(2 x+3)$
$+\mathrm{C}(x-1)(x+1) \quad \ldots(i)$
Put $\mathrm{x}=1,-1$ in (i), we get; $\mathrm{A}=-\frac{1}{10} \& \mathrm{~B}=\frac{5}{2}$
Putting $x=-\frac{3}{2}$ in $(\mathrm{i})$, we get : $\mathrm{C}=\frac{-24}{5}$
$\therefore \mathrm{I}=-\frac{1}{10} \int \frac{d x}{x-1}+\frac{5}{2} \int \frac{d x}{x+1}-\frac{24}{5} \int \frac{d x}{2 x+3}$
$=\frac{5}{2} \log |x+1|-\frac{1}{10} \log |x-1|-\frac{12}{5} \log |2 x+3|+\mathrm{C}$
$\equiv \frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{2 x+3}$
$\Rightarrow 2 x-3=\mathrm{A}(x+1)(2 x+3)+\mathrm{B}(x-1)(2 x+3)$
$+\mathrm{C}(x-1)(x+1) \quad \ldots(i)$
Put $\mathrm{x}=1,-1$ in (i), we get; $\mathrm{A}=-\frac{1}{10} \& \mathrm{~B}=\frac{5}{2}$
Putting $x=-\frac{3}{2}$ in $(\mathrm{i})$, we get : $\mathrm{C}=\frac{-24}{5}$
$\therefore \mathrm{I}=-\frac{1}{10} \int \frac{d x}{x-1}+\frac{5}{2} \int \frac{d x}{x+1}-\frac{24}{5} \int \frac{d x}{2 x+3}$
$=\frac{5}{2} \log |x+1|-\frac{1}{10} \log |x-1|-\frac{12}{5} \log |2 x+3|+\mathrm{C}$
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