Question

`int  ((2logx + 3))/(x(3logx + 2)[(logx)^2 + 1])  "d"x`

Answer 1

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"`