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

Apr 17, 2018

$\int 3 x \ln \left({x}^{3}\right) \mathrm{dx} = \frac{9 {x}^{2} \ln x}{2} - \frac{9}{4} {x}^{2} + C$

#### Explanation:

We're going to want to get rid of the ${x}^{3}$. Recalling that $\ln \left({x}^{a}\right) = a \ln x , \ln \left({x}^{3}\right) = 3 \ln x ,$ and we get

$3 \left(3\right) \int x \ln x \mathrm{dx} = 9 \int x \ln x \mathrm{dx}$ factoring all constants outside.

Now, we'll make the following selections for Integration by Parts:

$u = \ln x$

$\mathrm{du} = {x}^{-} 1 \mathrm{dx}$

$\mathrm{dv} = x \mathrm{dx}$

$v = \int x \mathrm{dx} = \frac{1}{2} {x}^{2}$

So, applying the formula, we get

$u v - \int v \mathrm{du} = \frac{{x}^{2} \ln x}{2} - \frac{1}{2} \int {x}^{-} 1 {x}^{2} \mathrm{dx}$

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

Recall that we need to multiply through by the $9$ we factored out:

$9 \left[\frac{{x}^{2} \ln x}{2} - \frac{1}{4} {x}^{2} + C\right] = \frac{9 {x}^{2} \ln x}{2} - \frac{9}{4} {x}^{2} + C$

Thus,

$\int 3 x \ln \left({x}^{3}\right) \mathrm{dx} = \frac{9 {x}^{2} \ln x}{2} - \frac{9}{4} {x}^{2} + C$

Apr 17, 2018

$\int 3 x \ln {x}^{3} \mathrm{dx} = \frac{9 {x}^{2}}{4} \left(2 \ln x - 1\right) + C$

#### Explanation:

Using the properties of logarithms:

$\ln {x}^{3} = 3 \ln x$

so:

$\int 3 x \ln {x}^{3} \mathrm{dx} = 9 \int x \ln x \mathrm{dx}$

integrate now by parts:

$\int 3 x \ln {x}^{3} \mathrm{dx} = 9 \int \ln x d \left({x}^{2} / 2\right)$

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

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

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

$\int 3 x \ln {x}^{3} \mathrm{dx} = \frac{9 {x}^{2} \ln x}{2} - \frac{9 {x}^{2}}{4} + C$

$\int 3 x \ln {x}^{3} \mathrm{dx} = \frac{9 {x}^{2}}{4} \left(2 \ln x - 1\right) + C$