How do you integrate $e^-lnx$ ?

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Answer 1 · 2015-04-06T18:01:08

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:

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

How do you integrate e^-lnxe^-lnx?