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Question
Evaluate: `int ("3x" - 1)/("2x"^2 - "x" - 1)` dx
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Solution
Let I = `int ("3x" - 1)/("2x"^2 - "x" - 1)` dx
`= int ("3x" - 1)/(("x - 1")("2x + 1"))` dx
Let `(3"x" - 1)/(("x - 1")("2x" + 1)) = "A"/"x - 1" + "B"/"2x + 1"`
∴ 3x - 1 = A(2x + 1) + B(x - 1) ...(i)
Putting x = 1 in (i), we get
3(1) - 1 = A(2 + 1) + B(0)
∴ 2 = 3A
∴ A = `2/3`
Putting x = `- 1/2` in (i), we get
`3(- 1/2) - 1 = "A"(0) + "B"[- 1/2 - 1]`
∴ `- 5/2 = "B" (- 3/2)`
∴ B = `5/3`
∴ `(3"x" - 1)/(("x" - 1)("2x" + 1)) = (2/3)/("x - 1") + (5/3)/("2x + 1")`
∴ I = `int ((2/3)/("x - 1") + (5/3)/("2x" + 1))` dx
`= 2/3 int 1/("x - 1") "dx" + 5/3 int 1/("2x + 1")`dx
∴ I = `2/3 log |"x - 1"| + 5/3 (log |("2x" + 1)|)/2` + c
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