Question

How do you evaluate the integral `int 1/(x+x^2)dx` from 1 to `oo`?

Answer 1

`ln 2`. Partial fractions (or equivalent).

Explanation:

`1/(x(1+x))=1/x-1/(1+x)`
So the integral is:
#[ln x - ln (1+x)]_1^oo =[ln(x/(1+x))]_1^oo#.

Since `x/(1+x)to1` as `x to oo` the definite integral becomes `ln 1 - ln (1/2)=0+ln2=ln2`.

Notice that the meaning of a definite integral with `oo` in a bound means "the limit as `a to oo` of the same definite integral but with `a` in that bound. You may get a nonsense if you try to use `oo` as a number!