# How do you integrate e^-lnx?

Apr 6, 2015

Use these two facts:

$- \ln x = \ln {x}^{- 1} = \ln \left(\frac{1}{x}\right)$

${e}^{\ln} u = u$

So,
$\int {x}^{- \ln x} \mathrm{dx} = \int \frac{1}{x} \mathrm{dx} = \ln \left\mid x \right\mid + C$.

Since the original function involves $\ln x$, we are justified in assuming that $x > 0$ and writing:

$\int {x}^{- \ln x} \mathrm{dx} = \ln x + C$.