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

Dec 12, 2016

$I = \frac{3 {x}^{2}}{4} \left[2 \ln x - 1\right] + C$

#### Explanation:

need to use integration by parts.

formula is:$\text{ } \int u \frac{\mathrm{dv}}{\mathrm{dx}} \mathrm{dx} = u v - \int v \frac{\mathrm{du}}{\mathrm{dx}} \mathrm{dx}$

the choice of $\text{ "u" }$&$\text{ } v$ is vital.

if logs are involved then it is usual to take $u = \ln f \left(x\right)$ but try and simplify the expression using the laws of logs first if possible.

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

$u = \ln x \implies \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x}$

$\frac{\mathrm{dv}}{\mathrm{dx}} = 3 x \implies v = \frac{3 {x}^{2}}{2}$

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

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

$I = \frac{3 {x}^{2}}{2} \ln x - \frac{3 {x}^{2}}{4} + C$

$I = \frac{3 {x}^{2}}{4} \left[2 \ln x - 1\right] + C$