How do you evaluate the integral #int 1/(1-x)dx# from 0 to 1?

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Answer 1 · 2017-01-10T23:56:56

The integral is not convergent

The integral is not convergent.

As the integrand is not continuous for #x=1# we can evaluate the improper integral:

#int_0^1 (dx)/(1-x) = lim_(t->1^-) int_0^t (dx)/(1-x)#

Now we have that:

# int_0^t (dx)/(1-x) =- int_0^t (d(1-x))/(1-x) = -[ln abs (1-x)]_0^t = - ln abs (1-t)#

and

#lim_(t->1^-) -ln(1-t) = +oo#