# How do you find the integral of x(ln x)^3 dx?

Jun 30, 2015

This is of course Integration by Parts.

Let:
$u = {\ln}^{3} x$
$\mathrm{du} = \frac{3 {\ln}^{2} x}{x} \mathrm{dx}$
$\mathrm{dv} = x \mathrm{dx}$
$v = {x}^{2} / 2$

$u v - \int v \mathrm{du}$

$= {x}^{2} / 2 {\ln}^{3} x - \int {x}^{\cancel{2}} / 2 \cdot \frac{3 {\ln}^{2} x}{\cancel{x}} \mathrm{dx}$

$= \frac{{x}^{2} {\ln}^{3} x}{2} - \frac{3}{2} \int x {\ln}^{2} x \mathrm{dx}$

Repeat:
$u = {\ln}^{2} x$
$\mathrm{du} = \frac{2 \ln x}{x} \mathrm{dx}$
$\mathrm{dv} = x \mathrm{dx}$
$v = {x}^{2} / 2$

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

$= \frac{{x}^{2} {\ln}^{3} x}{2} - \frac{3}{2} \left(\frac{{x}^{2} {\ln}^{2} x}{2} - \int {x}^{\cancel{2}} / \cancel{2} \cdot \frac{\cancel{2} \ln x}{\cancel{x}} \mathrm{dx}\right)$

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

and repeat again:
$u = \ln x$
$\mathrm{du} = \frac{1}{x} \mathrm{dx}$
$\mathrm{dv} = x \mathrm{dx}$
$v = {x}^{2} / 2$

$= \frac{{x}^{2} {\ln}^{3} x}{2} - \frac{3}{4} {x}^{2} {\ln}^{2} x + \frac{3}{2} \left(\int x \ln x \mathrm{dx}\right)$

$= \frac{{x}^{2} {\ln}^{3} x}{2} - \frac{3}{4} {x}^{2} {\ln}^{2} x + \frac{3}{2} \left(\frac{{x}^{2} \ln x}{2} - \int {x}^{\cancel{2}} / 2 \cdot \frac{1}{\cancel{x}} \mathrm{dx}\right)$

$= \frac{{x}^{2} {\ln}^{3} x}{2} - \frac{3}{4} {x}^{2} {\ln}^{2} x + \frac{3}{2} \left(\frac{{x}^{2} \ln x}{2} - \int \frac{x}{2} \mathrm{dx}\right)$

$= \frac{1}{2} {x}^{2} {\ln}^{3} x - \frac{3}{4} {x}^{2} {\ln}^{2} x + \frac{3}{4} {x}^{2} \ln x - \frac{3}{4} \int x \mathrm{dx}$

$= \frac{1}{2} {x}^{2} {\ln}^{3} x - \frac{3}{4} {x}^{2} {\ln}^{2} x + \frac{3}{4} {x}^{2} \ln x - \frac{3}{8} {x}^{2}$

$= \textcolor{b l u e}{\frac{1}{8} {x}^{2} \left(4 {\ln}^{3} x - 6 {\ln}^{2} x + 6 \ln x - 3\right) + C}$

Jun 30, 2015

I found:
${x}^{2} / 2 {\ln}^{3} \left(x\right) - \frac{3}{4} {x}^{2} {\ln}^{2} \left(x\right) + \frac{3}{4} {x}^{2} \ln \left(x\right) - \frac{3}{8} {x}^{2} + c$

#### Explanation:

I started integrating by Substitution and the by Parts (three times):