How do you evaluate the integral int (lnx)^3dx?

Feb 16, 2017

$\int {\left(\ln x\right)}^{3} \mathrm{dx} = x \left({\left(\ln x\right)}^{3} - 3 {\left(\ln x\right)}^{2} + 6 \ln x - 6\right) + C$

Explanation:

Substitute:

$t = \ln x$

$x = {e}^{t}$

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

so:

$\int {\left(\ln x\right)}^{3} \mathrm{dx} = \int {t}^{3} {e}^{t} \mathrm{dt}$

Integrate now by parts:

$\int {t}^{3} {e}^{t} \mathrm{dt} = \int {t}^{3} d \left({e}^{t}\right) = {t}^{3} {e}^{t} - 3 \int {t}^{2} {e}^{t} \mathrm{dt}$

$\int {t}^{2} {e}^{t} \mathrm{dt} = \int {t}^{2} d \left({e}^{t}\right) = {t}^{2} {e}^{t} - 2 \int t {e}^{t} \mathrm{dt}$

$\int t {e}^{t} \mathrm{dt} = \int t d \left({e}^{t}\right) = t {e}^{t} - \int {e}^{t} \mathrm{dt} = t {e}^{t} - {e}^{t} + C$

Putting it all together:

$\int {t}^{3} {e}^{t} \mathrm{dt} = {t}^{3} {e}^{t} - 3 {t}^{2} {e}^{t} + 6 t {e}^{t} - 6 {e}^{t} + C = {e}^{t} \left({t}^{3} - 3 {t}^{2} + 6 t - 6\right) + C$

and undoing the substitution:

$\int {\left(\ln x\right)}^{3} \mathrm{dx} = x \left({\left(\ln x\right)}^{3} - 3 {\left(\ln x\right)}^{2} + 6 \ln x - 6\right) + C$