Integrate the rational function: frac{2x-3}{( | vedclass

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(N/A) We have $\frac{2x-3}{(x^2-1)(2x+3)} = \frac{2x-3}{(x+1)(x-1)(2x+3)}$.Let $\frac{2x-3}{(x+1)(x-1)(2x+3)} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{

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(N/A) We have $\frac{2x-3}{(x^2-1)(2x+3)} = \frac{2x-3}{(x+1)(x-1)(2x+3)}$.
Let $\frac{2x-3}{(x+1)(x-1)(2x+3)} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{C}{2x+3}$.
Then $2x-3 = A(x-1)(2x+3) + B(x+1)(2x+3) + C(x+1)(x-1)$.
Setting $x=1$: $2(1)-3 = B(1+1)(2(1)+3) \Rightarrow -1 = B(2)(5) \Rightarrow B = -\frac{1}{10}$.
Setting $x=-1$: $2(-1)-3 = A(-1-1)(2(-1)+3) \Rightarrow -5 = A(-2)(1) \Rightarrow A = \frac{5}{2}$.
Setting $x=-\frac{3}{2}$: $2(-\frac{3}{2})-3 = C(-\frac{3}{2}+1)(-\frac{3}{2}-1) \Rightarrow -6 = C(-\frac{1}{2})(-\frac{5}{2}) \Rightarrow -6 = C(\frac{5}{4}) \Rightarrow C = -\frac{24}{5}$.
Thus,$\int \frac{2x-3}{(x^2-1)(2x+3)} dx = \int \left( \frac{5}{2(x+1)} - \frac{1}{10(x-1)} - \frac{24}{5(2x+3)} \right) dx$.
$= \frac{5}{2} \log|x+1| - \frac{1}{10} \log|x-1| - \frac{24}{5} \cdot \frac{1}{2} \log|2x+3| + K$.
$= \frac{5}{2} \log|x+1| - \frac{1}{10} \log|x-1| - \frac{12}{5} \log|2x+3| + K$,where $K$ is the constant of integration.

Integrate the rational function: $\frac{2x-3}{(x^2-1)(2x+3)}$

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