Question

Evaluate : `∫((1 + logx))/(x(2 + logx)(3 + logx)`dx

Answer 1

Let I = `∫((1 + logx))/(x(2 + logx)(3 + logx)`dx

Put logx = t

`(1/x)dx = dt`

∴ I = `∫(1 + t)/((2 + t)(3 + t))dt` ....(i)

By partial fraction 

`(1 + t)/((2 + t)(3 + t)) = A/(2 + t) + B/(3 + t)`

1 + t = A(3 + t) + B(2 + t) .....(ii)

Put t = -2 in equation (ii)

1 + (-2) = A(3 - 2) + B[2 + (-2)]

-1 = A + 0

∴ A = -1

Put t = -3 in equation (ii)

1 + (-3) = A[3 + (-3)] + B[2 + (-3)]

-2 = 0 - B

∴ B = 2

Putting A = -1 , B = 2 we get

`(1 + t)/((2 + t)(3 + t)) = (-1)/(2 + t) + 2/(3 + t)`

∴ Equation (i) becomes

I = `∫((-1)/(2 + t) + 2/(3 + t))` dt

= `-∫dt/(2 + t) + 2∫dt/(3 + t)`

∴ I = -log|2 + t| + 2 log|3 + t| + C

= -log|2 + logx| + 2 log|3 + logx| + C

= `log|(3 + logx)^2/(2 + logx)|` + C