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

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Answer 1 · 2016-08-29T20:53:25

That integral does not converge.

$x/(1-x)$ 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.

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