# How do you integrate int ln x^3 dx  using integration by parts?

Apr 19, 2016

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

#### Explanation:

First, simplify the function using the rule $\ln \left({a}^{b}\right) = b \cdot \ln a$; this gives the simplified integral

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

The question then becomes, how do we integrate $\int \ln x \mathrm{dx}$ by parts?

Integration by parts takes the form

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

So, we have to determine what to let $u$ and $\mathrm{dv}$ equal within $\int \ln x \mathrm{dx}$. Our best bet is to let $u = \ln x$ and simply have $\mathrm{dv} = 1 \mathrm{dx}$, since there's not much else we can do.

Find the values of $\mathrm{du}$ and $v$, by differentiating and integrating respectively.

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

ul(dv=1dx)" "=>" "intdv=intdx" "=>" "ul(v=x

Plugging these into the integration by parts formula, not forgetting the $3$ tagging along multiplicatively, we have:

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

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

$= 3 x \ln x - 3 x + C$