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

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Solution

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

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

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

$2 x - 3 = (2 A + 2 B + c) x^{2} + (5 A + B) x + (3 A - 3 B + c)$

Compare,

$2 A + 2 B + c = 0$ .....(1)

$5 A + B = 2$ .......(2)

$3 A - 3 B - C = - 3$ .....(3)

solve (1), (2) and (3) get

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

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

Integrate both side.
$= \frac{- 1}{10} log | x - 1 | + \frac{5}{2} log | x + 1 | \frac{- 12}{5} log | 2 x + 3 | + c$

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