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