Question

What is `int_1^oo 1/(1+x )-1/xdx`?

Answer 1

`= ln (1/2)`

Explanation:

`int_1^oo 1/(1+x )-1/xdx`

`= [ ln(1+x )- ln x ]_1^oo`

`= [ ln((1+x )/x) ]_1^oo`

`= [ ln((1/x+1 )/1) ]_1^oo`

`= lim_{x to oo} ln((1/x+1 )/1) - ln((1/1+1 )/1)`

`= lim_{x to oo} ln((1/x+1 )/1) - ln((1/1+1 )/1)`

`= ln 1 - ln 2 = ln (1/2)`