# What is the antiderivative of ln(x)?

Mar 5, 2015

$I = \int \ln \left(x\right) \mathrm{dx}$

Let $\ln \left(x\right) = t$

$\implies x = {e}^{t}$

$\implies \mathrm{dx} = {e}^{t} \mathrm{dt}$

Substituting in the Integral,

$I = \int t {e}^{t} \mathrm{dt}$

On integrating by parts, keeping the first function as $t$ and second function as ${e}^{t}$, we get

$I = t \int {e}^{t} \mathrm{dt} - \int \left(\frac{\mathrm{dt}}{\mathrm{dt}} \int {e}^{t} \mathrm{dt}\right) \mathrm{dt}$

Which is, simply,

$I = t {e}^{t} - {e}^{t} + C$

$\implies I = {e}^{t} \left(t - 1\right) + C$

Substituting the value of $t = \ln \left(x\right)$,

$\int \ln \left(x\right) \mathrm{dx} = x \left[\ln \left(x\right) - 1\right] + C$