# Evaluate the integral? :  int_1^e x^5 (lnx)^2 dx

May 31, 2017

${\int}_{1}^{e} \setminus {x}^{5} \setminus {\left(\ln x\right)}^{2} \setminus \mathrm{dx} = \frac{13 {e}^{6} - 1}{108}$

#### Explanation:

We want to find:

$I = {\int}_{1}^{e} \setminus {x}^{5} \setminus {\left(\ln x\right)}^{2} \setminus \mathrm{dx}$

We can use integration by Parts

Let $\left\{\begin{matrix}u & = {\left(\ln x\right)}^{2} & \implies \frac{\mathrm{du}}{\mathrm{dx}} & = \frac{2 \ln x}{x} \\ \frac{\mathrm{dv}}{\mathrm{dx}} & = {x}^{5} & \implies v & = {x}^{6} / 6\end{matrix}\right.$

Then plugging into the IBP formula:

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

gives us

${\int}_{1}^{e} \setminus {\left(\ln x\right)}^{2} \left({x}^{5}\right) \setminus \mathrm{dx} = {\left[{\left(\ln x\right)}^{2} \left({x}^{6} / 6\right)\right]}_{1}^{e} - {\int}_{1}^{e} \setminus \left({x}^{6} / 6\right) \left(\frac{2 \ln x}{x}\right) \setminus \mathrm{dx}$

$\therefore I = \frac{1}{6} \left({e}^{6} {\ln}^{2} e - {\ln}^{21}\right) - \frac{1}{3} \setminus {\int}_{1}^{e} \setminus {x}^{5} \ln x \setminus \mathrm{dx}$
$\text{ } = {e}^{6} / 6 - \frac{1}{3} \setminus {\int}_{1}^{e} \setminus {x}^{5} \ln x \setminus \mathrm{dx}$

So now let us look at this next integral:

$J = {\int}_{1}^{e} \setminus {x}^{5} \ln x \setminus \mathrm{dx}$

Again we can use integration by Parts

Let $\left\{\begin{matrix}u & = \ln x & \implies \frac{\mathrm{du}}{\mathrm{dx}} & = \frac{1}{x} \\ \frac{\mathrm{dv}}{\mathrm{dx}} & = {x}^{5} & \implies v & = {x}^{6} / 6\end{matrix}\right.$

and plugging into the IBP formula gives us

${\int}_{1}^{e} \setminus \left(\ln x\right) \left({x}^{5}\right) \setminus \mathrm{dx} = {\left[\left(\ln x\right) \left({x}^{6} / 6\right)\right]}_{1}^{e} - {\int}_{1}^{e} \setminus \left({x}^{6} / 6\right) \left(\frac{1}{x}\right) \setminus \mathrm{dx}$

$\therefore J = \frac{1}{6} \left({e}^{6} \ln e - \ln 1\right) - \frac{1}{6} \setminus {\int}_{1}^{e} \setminus {x}^{5} \setminus \mathrm{dx}$
$\text{ } = {e}^{6} / 6 - \frac{1}{6} \setminus {\left[{x}^{6} / 6\right]}_{1}^{e}$
$\text{ } = {e}^{6} / 6 - \frac{1}{36} \setminus \left({e}^{6} - 1\right)$

Combining these results we get:

$I = {e}^{6} / 6 - \frac{1}{3} \left({e}^{6} / 6 - \frac{1}{36} \setminus \left({e}^{6} - 1\right)\right)$
$\setminus \setminus = {e}^{6} / 6 - \frac{1}{3} \left({e}^{6} / 6 - {e}^{6} / 36 + \frac{1}{36}\right)$
$\setminus \setminus = {e}^{6} / 6 - \frac{1}{3} \left(\frac{5}{36} {e}^{6} + \frac{1}{36}\right)$
$\setminus \setminus = {e}^{6} / 6 - \frac{5}{108} {e}^{6} - \frac{1}{108}$
$\setminus \setminus = \frac{13}{108} {e}^{6} - \frac{1}{108}$
$\setminus \setminus = \frac{13 {e}^{6} - 1}{108}$