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Question
Evaluate the following:
`int (2x - 1)/((x - 1)(x + 2)(x - 3)) "d"x`
Solution
Let I = `int (2x - 1)/((x - 1)(x + 2)(x - 3)) "d"x`
Resolving into partial fraction, we put
`(2x - 1)/((x - 1)(x + 2)(x - 3)) = "A"/(x - 1) + "B"/(x + 2) + "C"/(x - 3)`
⇒ 2x – 1 = A(x + 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x + 2)
Put x = 1
1 = A(3)(– 2)
⇒ A = `-1/6`
Put x = – 2
– 5 = B(– 3)(– 5)
⇒ B = `- 1/3`
Put x = 3
5 = C(2)(5)
⇒ C = `1/2`
∴ `int (2x - 1)/((x - 1)(x + 2)(x - 3)) "d"x = - 1/6 int 1/(x - 1) "d"x - 1/3 int 1/(x + 2) "d"x + 1/2 int 1/(x - 3) "d"x`
= `- 1/6 log |x - 1| - 1/3 log|x + 2| + 1/2 log|x - 3| + "C"`
= `- log|x - 1|^(1/6) - log(x + 2)^(1/3) + log(x - 3)^(1/3) + "C"`
Hence, `int (2x - 1)/((x - 1)(x + 2)(x - 3)) "d"x = log[sqrt(x - 3)/((x - 1)^(1/6) (x + 2)^(1/3))] + "C"`
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