# How do you integrate int6x ln x dx from 2 to 4?

Mar 18, 2018

${\int}_{2}^{4} \setminus 6 x \ln x \setminus \mathrm{dx} = 84 \ln 2 - 18 \approx 40.224 \setminus \left(3 \setminus \mathrm{dp}\right)$

#### Explanation:

We seek:

$I = {\int}_{2}^{4} \setminus 6 x \ln x \setminus \mathrm{dx}$

We can then apply Integration By Parts:

Let  { (u,=lnx, => (du)/dx,=1/x), ((dv)/dx,=6x, => v,=3x^2 ) :}

Then plugging into the IBP formula:

$\int \setminus \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \setminus \mathrm{dx} = \left(u\right) \left(v\right) - \int \setminus \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \setminus \mathrm{dx}$

We have:

${\int}_{2}^{4} \left(\ln x\right) \left(6 x\right) \setminus \mathrm{dx} = {\left[\left(\ln x\right) \left(3 {x}^{2}\right)\right]}_{2}^{4} - {\int}_{2}^{4} \setminus \left(3 {x}^{2}\right) \left(\frac{1}{x}\right) \setminus \mathrm{dx}$

$\therefore I = {\left[3 {x}^{2} \ln x\right]}_{2}^{4} - {\int}_{2}^{4} \setminus 3 x \setminus \mathrm{dx}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 3 \left\{16 \ln 4 - 4 \ln 2\right\} - {\left[\frac{3 {x}^{2}}{2}\right]}_{2}^{4}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 3 \left\{16 \ln {2}^{2} - 4 \ln 2\right\} - \frac{3}{2} {\left[{x}^{2}\right]}_{2}^{4}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 3 \left\{32 \ln 2 - 4 \ln 2\right\} - \frac{3}{2} \left\{16 - 4\right\}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 3 \left\{28 \ln 2\right\} - \frac{3}{2} \cdot 12$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 84 \ln 2 - 18$