# ∫ (2logx+3)x(3logx+2)[(logx)2+1] dx - Mathematics and Statistics

Sum

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

#### Solution

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"

Is there an error in this question or solution?
Chapter 2.3: Indefinite Integration - Long Answers III

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