Question

How do you evaluate the integral `int x/(1-x)dx` from 0 to 2?

Answer 1

That integral does not converge.

Explanation:

`x/(1-x)` is undefined at `x=1`, so the integral is improper. We have to break it into

`int_0^1 x/(1-x) dx + int_1^2 x/(1-x) dx` which converges if the two integrals converge.

`int_0^1 x/(1-x) dx = lim_(brarr1^-) int_0^b (x-1+1)/(1-x) dx`

`= lim_(brarr1^-) int_0^b (-1 +1/(1-x)) dx`

`= lim_(brarr1^-) (-x-ln abs(1-x))]_0^b`

`= lim_(brarr1^-) (-b-ln(1-b)) - (-0-ln1)`

But `lim_(brarr1^-)(-1-ln(1-b)) = oo`.

So the integral diverges.