# How do you integrate (ln x / 3)^3?

Apr 20, 2016

$\int {\left(\ln \frac{x}{3}\right)}^{3} \mathrm{dx} = \frac{x {\ln}^{3} x}{27} - \frac{x {\ln}^{2} x}{9} + \frac{2 x \ln x}{9} - \frac{2 x}{9} + C$

#### Explanation:

Note that ${\left(\ln \frac{x}{3}\right)}^{3} = {\ln}^{3} \frac{x}{27}$. From this, we see that

$\int {\left(\ln \frac{x}{3}\right)}^{3} \mathrm{dx} = \frac{1}{27} \int {\ln}^{3} x \mathrm{dx}$

Using integration by parts:

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

We let

$u = {\ln}^{3} x \text{ "=>" } \mathrm{du} = \frac{3 {\ln}^{2} x}{x} \mathrm{dx}$

$\mathrm{dv} = \left(1\right) \mathrm{dx} \text{ "=>" } v = x$

This gives us:

$\frac{1}{27} \int {\ln}^{3} x \mathrm{dx} = \frac{1}{27} x {\ln}^{3} x - \frac{1}{27} \int 3 {\ln}^{2} x \mathrm{dx}$

$= \frac{1}{27} x {\ln}^{3} x - \frac{1}{9} \int {\ln}^{2} x \mathrm{dx}$

Integrate $\int {\ln}^{2} x \mathrm{dx}$ similarly (using by parts again):

$u = {\ln}^{2} x \text{ "=>" } \mathrm{du} = \frac{2 \ln x}{x} \mathrm{dx}$

$\mathrm{dv} = \left(1\right) \mathrm{dx} \text{ "=>" } v = x$

Thus,

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

Combining this and multiplying by $- \frac{1}{9}$, we see that:

$\frac{1}{27} \int {\ln}^{3} x \mathrm{dx} = \frac{x {\ln}^{3} x}{27} - \frac{x {\ln}^{2} x}{9} + \frac{2}{9} \int \ln x \mathrm{dx}$

Use integration by parts one last time:

$u = \ln x \text{ "=>" } \mathrm{du} = \frac{1}{x} \mathrm{dx}$

$\mathrm{dv} = \left(1\right) \mathrm{dx} \text{ "=>" } v = x$

Thus,

$\int \ln x \mathrm{dx} = x \ln x - \int \mathrm{dx} = x \ln x - x$

Hence,

$\frac{1}{27} \int {\ln}^{3} x \mathrm{dx} = \frac{x {\ln}^{3} x}{27} - \frac{x {\ln}^{2} x}{9} + \frac{2 x \ln x}{9} - \frac{2 x}{9} + C$

Don't forget the constant of integration!