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प्रश्न
`int ((2logx + 3))/(x(3logx + 2)[(logx)^2 + 1]) "d"x`
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उत्तर
Let I = `int ((2logx + 3))/(x(3logx + 2)[(logx)^2 + 1]) "d"x`
Put log x = t
∴ `1/x "d"x = dt`
∴ I = `int (2"t" + 3)/((3"t" + 2)("t"^2 + 1)) "dt"`
Let `(2 + 3)/((3"t" + 2)("t"^2 + 1)) = "A"/(3"t" + 2) + ("Bt" + "C")/("t"^2 + 1)`
∴ 2t + 3 = A(t2 + 1) + (Bt + C)(3t + 2) .........(i)
Putting t = `-2/3` in (i), we get
`2((-2)/3) + 3 = "A"[((-2)/3)^2 + 1]`
∴ `(-4)/3 + 3 = "A"(4/9 + 1)`
∴ `5/3 = "A"(13/9)`
∴ A = `15/13`
Putting t = 0 in (i), we get
3 = A(1) + C(2)
∴ 3 = `15/13 + 2"C"`
∴ `3 - 15/13` = 2C
∴ `24/13` = 2C
∴ C = `12/13`
Putting t = 1 in (i), we get
2 + 3 = A(1 + 1) + (B + C)(3 + 2)
∴ 5 = 2A + 5(B + C)
∴ 5 = `2(15/13) + 5("B" + 12/13)`
∴ 5 = `30/13 + 5"B" + 60/13`
∴ 5B = `5 - 30/13 - 60/13`
∴ 5B = `-25/13`
∴ B = `(-5)/13`
∴ `(2"t" + 3)/((3"t" + 2)("t"^2 + 1)) = (15/13)/(3"t" + 2) + (-5/13 "t" + 12/13)/("t"^2 + 1)`
∴ I = `int((15/13)/(3"t" + 2) + ((-5)/13 "t" + 12/13)/("t"^2 + 1)) "dt"`
= `15/13 int 1/(3"t" + 2) "dt" - 5/13 int "t"/("t"^2 + 1) "dt" + 12/13 int 1/("t"^2 + 1) "dt"`
= `15/13 int 1/(3"t" + 2) "dt" - 5/13*1/2 int (2"t")/("t"^2 + 1) "dt" + 12/13 int 1/("t"^2 + 1) "dt"`
= `15/13* (log|3"t" + 2|)/3 - 5/26 log|"t"^2 + 1| + 12/13 tan^-1 "t" + "c"`
∴ I = `5/13 log |3 log x 2| - 5/26 log |(logx)^2 + 1| + 12/13 tan^-1(logx) + "c"`
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