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Question
Integrate the rational function:
`(2x - 3)/((x^2 -1)(2x + 3))`
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Solution
Let `(2x - 3)/((x^2 - 1)(2x + 3))`
`= (2x - 3)/((x - 1)(x + 1) (2x + 3))`
`= A/(x - 1) + B/(x + 1) + C/(2x + 3)`
⇒ 2x - 3 = A(x + 1)(2x + 3) + B(x - 1)(2x + 3) + C(x - 1)(x + 1) .... (1)
Putting x = 1 in equation (1),
2(1) - 3 = A(1 + 1)(2 + 3)
⇒ -1 = A (2) (5)
⇒ A `= -1/10`
Putting x = -1 in equation (1),
-2 -3 = B (-1 -1)(-2 + 3)
⇒ -5 = B (-2)(1)
⇒ B `= 5/2`
Putting `x = -3/2` in equation (1),
-3 -3 = C `(-3/2 -1)(-3/2 + 1)`
⇒ -6 = C `(-5/2)(-1/2)`
⇒ C =`- 6 xx 4/5 = -24/5`
`therefore (2x - 3)/((x^2 - 1)(2x + 3)) = - 1/(10(x - 1)) + 5/(2(x + 1)) - 24/(5(2x + 3))`
`therefore int (2x - 3)/((x^2 - 1)(2x+ 3)) dx = -1/10 int 1/(x - 1) dx + 5/2 int 1/(x + 1) dx -24/5 int 1/(2x + 3) dx`
` = - 1/10 log (x - 1) + 5/2 log (x + 1) - 24/5 log ((2x + 3)/2) + C`
`= 5/2 log (x + 1) - 1/10 log (x - 1) - 12/5 log (2x+ 3) + C`
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