Question

How do you integrate `e^-lnx`?

Answer 1

Use these two facts:

`-lnx = ln x^(-1)=ln(1/x)`

`e^lnu = u`

So,
`int x^(-lnx) dx = int 1/x dx=ln absx+C`.

Since the original function involves `lnx`, we are justified in assuming that `x>0` and writing:

`int x^(-lnx) dx =lnx+C`.