# How do you integrate 6x ln x dx ?

Apr 13, 2015

Integrate by parts.

$\int 6 x \ln x \mathrm{dx}$

Let $u = \ln x$ so $\mathrm{du} = \frac{1}{x} \mathrm{dx}$ and

let $\mathrm{dv} = 6 x \mathrm{dx}$ to get $v = 3 {x}^{2}$

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

$\int 6 x \ln x \mathrm{dx} = 3 {x}^{2} \ln x - \int 3 {x}^{2} \frac{1}{x} \mathrm{dx}$

$\textcolor{w h i t e}{\text{ssssssssssss}}$ $= 3 {x}^{2} \ln x - \int 3 x \mathrm{dx}$

$\textcolor{w h i t e}{\text{ssssssssssss}}$ $= 3 {x}^{2} \ln x - \frac{3}{2} {x}^{2} + C$

Apr 13, 2015

The answer is: $3 {x}^{2} \ln x - 3 {x}^{2} / 2 + c$.

This integral has to be done with the theorem of the integration by parts, that says:

$\int f \left(x\right) g ' \left(x\right) \mathrm{dx} = f \left(x\right) g \left(x\right) - \int g \left(x\right) f ' \left(x\right) \mathrm{dx}$

We can assume that $f \left(x\right) = \ln x$ and $g ' \left(x\right) \mathrm{dx} = 6 x \mathrm{dx}$.

$g \left(x\right) = 6 {x}^{2} / 2 = 3 {x}^{2}$

$f ' \left(x\right) \mathrm{dx} = \frac{1}{x} \mathrm{dx}$.

So:

$\int 6 x \ln x \mathrm{dx} = 3 {x}^{2} \ln x - \int 3 {x}^{2} \cdot \frac{1}{x} \mathrm{dx} = 3 {x}^{2} \ln x - 3 \int x \mathrm{dx} =$

$= 3 {x}^{2} \ln x - 3 {x}^{2} / 2 + c$.

Apr 13, 2015

$= 3 {x}^{2} \ln x - \frac{1}{4} {x}^{2} + C$

Detail

$\int 6 x \ln x \mathrm{dx}$

$= \ln x \frac{6 {\left(x\right)}^{2}}{2} - \int \frac{1}{x} . \frac{6 {x}^{2}}{2} \mathrm{dx}$

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

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

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