# How do you integrate (t^2)ln(t) using integration by parts?

## ${\int}_{1}^{{e}^{2}} {t}^{2} \ln \left(t\right) \mathrm{dt}$

Jan 24, 2018

$\frac{1}{9} \left(5 \cdot {e}^{6} + 1\right)$

#### Explanation:

We want to integrate

${\int}_{1}^{{e}^{2}} {t}^{2} \cdot \ln \left(t\right) \mathrm{dt}$

Use integration by parts $\int u \mathrm{dv} = u v - \int v \mathrm{du}$

Let $u = \ln \left(t\right)$ and $\mathrm{dv} = {t}^{2} \mathrm{dt}$

Then $\mathrm{du} = \frac{1}{t} \mathrm{dt}$ and $v = \frac{1}{3} {t}^{3}$

${\left[\frac{1}{3} \ln \left(t\right) \cdot {t}^{3}\right]}_{1}^{{e}^{2}} - \frac{1}{3} {\int}_{1}^{{e}^{2}} {t}^{2} \mathrm{dt}$

${\left[\frac{1}{3} \ln \left(t\right) \cdot {t}^{3}\right]}_{1}^{{e}^{2}} - {\left[\frac{1}{9} {t}^{3}\right]}_{1}^{{e}^{2}}$

$\frac{1}{3} \ln \left({e}^{2}\right) \cdot {\left({e}^{2}\right)}^{3} - \frac{1}{9} {\left({e}^{2}\right)}^{3} + \frac{1}{9} \cdot {1}^{3}$

$\frac{6}{9} \cdot {e}^{6} - \frac{1}{9} {e}^{6} + \frac{1}{9}$

$\frac{1}{9} \left(5 \cdot {e}^{6} + 1\right)$

Jan 24, 2018

The answer is $= 224.2$

#### Explanation:

$\int u ' v = u v - \int u v '$

Here,

intt^2lntdt=?

Let $v = \ln t$, $\implies$, $v ' = \frac{1}{t}$

$u ' = {t}^{2}$, $\implies$, $u = {t}^{3} / 3$

Therefore,

$\int {t}^{2} \ln t \mathrm{dt} = {t}^{3} / 3 \ln t - \int \frac{1}{t} \cdot {t}^{3} / 3 \mathrm{dt}$

$= {t}^{3} / 3 \ln t - \frac{1}{3} \int {t}^{2} \mathrm{dt}$

$= {t}^{3} / 3 \ln t - \frac{1}{9} {t}^{3} + C$

Then, the definite integral is

${\int}_{1}^{{e}^{2}} {t}^{2} \ln t \mathrm{dt} = {\left[{t}^{3} / 3 \ln t - \frac{1}{9} {t}^{3}\right]}_{1}^{{e}^{2}}$

$= \left({e}^{6} / 3 \ln \left({e}^{2}\right) - {e}^{6} / 9\right) - \left(\frac{1}{3} \ln \left(1\right) - \frac{1}{9}\right)$

$= \frac{5}{9} {e}^{6} + \frac{1}{9}$

$= 224.2$

Jan 24, 2018

${\int}_{1}^{{e}^{2}} {t}^{2} \ln \left(t\right) \mathrm{dt} = \frac{5 {e}^{6} + 1}{9} \approx 224.23822$

#### Explanation:

First solve for the indefinite integral

Using integration by parts

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

Let

$u = \ln t \implies \mathrm{du} = \frac{1}{t} \mathrm{dt}$

$\mathrm{dv} = {t}^{2} \implies v = {t}^{3} / 3 \leftarrow$ (See Proof Below)

Use color(blue)(intt^adt=(t^(a+1))/(a+1)

So color(red)(intt^2dt=t^(2+1)/(2+1)=t^3/3

Substituting in for the formula:

$I = \left(\ln \left(t\right)\right) \cdot \left({t}^{3} / 3\right) - \int \left({t}^{3} / 3\right) \left(\frac{1}{t} \mathrm{dt}\right)$

Simplify (Take out the constant $\frac{1}{3}$):

$I = \frac{{t}^{3} \ln \left(t\right)}{3} - \frac{1}{3} \int {t}^{3} / t \mathrm{dt}$

$I = \frac{{t}^{3} \ln \left(t\right)}{3} - \frac{1}{3} \int {t}^{2} \mathrm{dt}$

Now we solve for $\int {t}^{2} \mathrm{dt}$ but actually solved this earlier at the start of the problem:

So...

$I = \frac{{t}^{3} \ln \left(t\right)}{3} - \frac{1}{3} \left[{t}^{3} / 3\right]$

Simplify

$I = {\left[\frac{{t}^{3} \ln \left(t\right)}{3} - {t}^{3} / 9\right]}_{1}^{{e}^{2}}$

Now we evaluate the upper/lower limits

$I = \left[\frac{{\textcolor{red}{\left({e}^{2}\right)}}^{3} \ln \left(\textcolor{red}{{e}^{2}}\right)}{3} - {\left(\textcolor{red}{{e}^{2}}\right)}^{3} / 9\right] - \left[\frac{{\textcolor{red}{1}}^{3} \ln \left(\textcolor{red}{1}\right)}{3} - {\textcolor{red}{1}}^{3} / 9\right]$

$I = \frac{5 {e}^{6}}{9} - \left(- \frac{1}{9}\right)$

$I = \frac{5 {e}^{6}}{9} + \frac{1}{9}$

$I = \frac{5 {e}^{6} + 1}{9} \approx 224.23822$