Evaluate: ∫3x - 2(x + 1)2(x + 3) dx - Mathematics and Statistics

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Sum

Evaluate: `int "3x - 2"/(("x + 1")^2("x + 3"))` dx

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Solution

Let I = `int "3x - 2"/(("x + 1")^2("x + 3"))` dx

Let `"3x - 2"/(("x + 1")^2("x + 3")) = "A"/"x + 1" + "B"/("x + 1")^2 + "C"/("x + 3")`

∴ 3x - 2 = (x + 3) [A(x + 1) + B] + C(x + 1)2  ....(i)

Putting x = - 1 in (i), we get 

3(- 1) - 2 = (–1 + 3)[A(0) + B] + C(0)

∴ - 5 = 2B

∴ B = -`5/2`

Putting x = - 3 in (i), we get

3(- 3)-2 = 0[A(–3 + 1) + B] + C(–2)2

∴ - 11 = 4C

∴ C = - `11/4`

Putting x = 0 in (i), we get

3(0)- 2 = 3[A(0 + 1) + B] + C(0 + 1)2

∴ - 2 = 3A + 3B + C

∴ - 2 = 3A + 3`(- 5/2) - 11/4`

∴ 3A = –2 + `15/2 + 11/4 = (- 8 + 30  11)/4 = 33/4`

∴ A = `33/4 xx 1/3 = 11/4`

∴ `"3x - 2"/(("x + 1")^2("x + 3")) = (11/4)/"x + 1" + (- 5/2)/("x + 1")^2 + (- 11/4)/"x + 3"`

∴ I = `int ((11/4)/"x + 1" - (5/2)/("x + 1")^2 - ( 11/4)/"x + 3")` dx

`= 11/4 int "dx"/"x + 1" - 5/2 int ("x + 1")^-2 "dx" - 11/4 int "dx"/"x + 3"`

`= 11/4 log |"x + 1"| - 5/2 (- 1/"x + 1") - 11/4 log |"x + 3"|` + c

`= 11/4 [log |"x" + 1| - log |"x" + 3|] + 5/(2("x" + 1))` + c

∴ I = `11/4 log |("x + 1")/("x + 3")| + 5/(2("x + 1"))` + c

  Is there an error in this question or solution?
Chapter 5: Integration - Exercise 5.6 [Page 135]

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