Question

Integrate the following w.r.t. x : `(2log x + 3)/(x(3 log x + 2)[(logx)^2 + 1]`

Answer 1

Let I = `int (2log x + 3)/(x(3 log x + 2)[(logx)^2 + 1]]*dx`

Put log x = t
∴ `(1)/x*dx` = dt

∴ I = `int (2t + 3)/((3t + 2)(t^2 + 1))*dt`

Let `(2t + 3)/((3t + 2)(t^2 + 1)) = "A"/(3t + 2) + "Bt + C"/(t^2 + 1)`

∴ 2t + 3 = A(t² + 1) + (Bt + C)(3t + 2)

Put 3t + 2 = 0 i,e, t = `-(2)/(3)`, we get

`2((-2)/3) + 3 = "A"(4/9 + 1) + ((-2)/3 "B" + "C")(0)`

∴ `(5)/(3) = "A"(13/9)`

∴ A = `(15)/(13)`
Put t = 0, we get

3 = A(1) + C(2) = `(15)/(13) + 2"C"`

∴ 2C = `3 - (15)/(13) = (24)/(13)`

∴ C = `(12)/(13)`
Comparing coefficient of t² on both the sides, we get
0 = A + 3B

∴ B = `- "A"/(3) = - (5)/(13)`

∴ `(2t + 3)/((3t + 2)(t^2 + 1)) = ((15/13))/(3t + 2) + ((-5/13t + 2/13))/(t^2 + 1)`

∴ I = `int [((15/13))/(3t + 2) + ((-5/13t + 12/3))/(t^2 + 1)]*dt`

= `(15)/(13) int 1/(3t + 2)*dt - (5)/(26) int (2t)/(t*^2 + 1)*dt + (12)/(13) int 1/(t^2 + 1)*dt`

= `(15)/(13)*(1)/(3)log|3t + 2| - (5)/(26)log|t^2 + 1| + (12)/(13)tan^-1 (t) + c`

...`[because d/dt (t^2 + 1) = 2t and int (f'(x))/f(x)dt = log|f(t)| + c]`

= `(5)/(13)log|3logx + 2| - (5)/(26)log|(logx)^2 + 1| + (12)/(13)tan^-1(logx) + c`.