How to integrate (x-1)/logx w.r.t. x in the interval (0,1)?

Question

this is a definete integral with lower limit =0, upper limit =1
log implies to the base 'e'

Answer 1

Ahh Santmann if u cant solve the question then accept it. Dont talk nonsense. This problem isn't wrong. I will demonstrate.

Consider
f(t) = int(0 to1, (x^t-1)/logx dx)
f'(t) = int(0 to1, (x^t)logx/logx dx)... diff. under integral sign
= int(0 to 1, x^t dx)
= int(0 to 1, [x^(t+1)]/(t+1))
= 1/(t+1)
f(t) = log(t+1) + c
f(0) = int(0 to1, (x^0-1)/logx dx)
=0
f(0) = log(0+1) +c = c
c = f(0) = 0
f(t) = log(t+1)
Barad what u want is f(1)
so f(1) = log2
ur ans is log2

Answer 2

In the first place this is an improper integral bcause log x does not exist at 0 and its limit i minus infinite.
On the other side,as far as I know the primitive of this function is NOT expressable by elementary functions.
I think there must be some mistake in writing

Answer 3

you may write it as quintessential of logx*(a million/x)dx. i'm assuming its a organic log. now as u can see by-fabricated from logx (i.e a million/x) is invloved in quintessential subsequently u could be conscious ability rule of integration. answer is quintessential of logx*a million/x*dx ------(A) substitue t=logx now dt=a million/x*dx now fact A will appear as if this quintessential of t*dt on integrating you will get t^2 + C now putting back the fee of 't' will yield you ((logx)^2)/2 +c n tht's ur answer!!!!