# What is the antiderivative of (ln(x))^3?

Jan 8, 2016

I found: $x {\ln}^{3} \left(x\right) - 3 x {\ln}^{2} \left(x\right) + 6 x \ln \left(x\right) - 6 x + c$

#### Explanation:

We can try to evaluate:
$\int {\left(\ln \left(x\right)\right)}^{3} \mathrm{dx} =$
Set $\ln \left(x\right) = t$
$x = {e}^{t}$
$\mathrm{dx} = {e}^{t} \mathrm{dt}$
So we get:
$= \int {t}^{3} {e}^{t} \mathrm{dt} =$ we can try By Parts:
$= {t}^{3} {e}^{t} - \int 3 {t}^{2} {e}^{t} \mathrm{dt} = {t}^{3} {e}^{t} - 3 \int {t}^{2} {e}^{t} \mathrm{dt} =$
Again:
$= {t}^{3} {e}^{t} - 3 \left[{t}^{2} {e}^{t} - \int 2 t {e}^{t} \mathrm{dt}\right] =$
$= {t}^{3} {e}^{t} - 3 {t}^{2} {e}^{t} + 6 \int t {e}^{t} \mathrm{dt} =$
Again:
$= {t}^{3} {e}^{t} - 3 {t}^{2} {e}^{t} + 6 t {e}^{t} - 6 \int {e}^{t} \mathrm{dt} =$
$= {t}^{3} {e}^{t} - 3 {t}^{2} {e}^{t} + 6 t {e}^{t} - 6 {e}^{t} + c$
But $t = \ln \left(x\right)$
And remember that ${e}^{\ln \left(x\right)} = x$. So:
$= x {\ln}^{3} \left(x\right) - 3 x {\ln}^{2} \left(x\right) + 6 x \ln \left(x\right) - 6 x + c$