# How do you integrate int x^3lnx by integration by parts method?

Oct 25, 2016

$\int {x}^{3} \ln \left(x\right) \mathrm{dx} = \frac{{x}^{4} \left(4 \ln \left(x\right) - 1\right)}{16} + C$

#### Explanation:

Integration by parts takes the form:

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

So, for the integral $\int {x}^{3} \ln \left(x\right) \mathrm{dx}$, we see this is $\int u \mathrm{dv}$ and let:

$\left\{\begin{matrix}u = \ln \left(x\right) \text{ "=>" "du=1/xdx \\ dv=x^3dx" "=>" } v = {x}^{4} / 4\end{matrix}\right.$

Thus:

$\int {x}^{3} \ln \left(x\right) \mathrm{dx} = u v - \int v \mathrm{du} = \frac{{x}^{4} \ln \left(x\right)}{4} - \int {x}^{4} / 4 \frac{1}{x} \mathrm{dx}$

Simplifying the integral:

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

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

$\int {x}^{3} \ln \left(x\right) \mathrm{dx} = \frac{{x}^{4} \ln \left(x\right)}{4} - {x}^{4} / 16 + C$

$\int {x}^{3} \ln \left(x\right) \mathrm{dx} = \frac{{x}^{4} \left(4 \ln \left(x\right) - 1\right)}{16} + C$

Oct 25, 2016

$\int {x}^{3} \ln x \mathrm{dx} = \frac{1}{4} {x}^{4} \ln x - {x}^{4} / 16 + C$

#### Explanation:

The formula for integration by parts is:
$\int u \frac{\mathrm{dv}}{\mathrm{dx}} \mathrm{dx} = u v - \int v \frac{\mathrm{du}}{\mathrm{dx}} \mathrm{dx}$

Essentially we would like to find one function that simplifies when differentiated, and one that simplifies when integrated (or is at least integrable).

Let $\left\{\begin{matrix}u = \ln x & \implies & \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x} \\ \frac{\mathrm{dv}}{\mathrm{dx}} = {x}^{3} & \implies & v = {x}^{4} / 4\end{matrix}\right.$

So IBP gives:

$\int \left(\ln x\right) \left({x}^{3}\right) \mathrm{dx} = \left(\ln x\right) \left({x}^{4} / 4\right) - \int \left({x}^{4} / 4\right) \left(\frac{1}{x}\right) \mathrm{dx}$
$\therefore \int {x}^{3} \ln x \mathrm{dx} = \frac{1}{4} {x}^{4} \ln x - \frac{1}{4} \int {x}^{3} \mathrm{dx}$
$\therefore \int {x}^{3} \ln x \mathrm{dx} = \frac{1}{4} {x}^{4} \ln x - \frac{1}{4} \left({x}^{4} / 4\right) + C$
$\therefore \int {x}^{3} \ln x \mathrm{dx} = \frac{1}{4} {x}^{4} \ln x - {x}^{4} / 16 + C$