# How do you integrate int x^3lnabsx from 1 to 2 by integration by parts method?

Aug 24, 2016

$= 4 \ln 2 - \frac{15}{16}$

#### Explanation:

${\int}_{1}^{2} {x}^{3} \ln \left\mid x \right\mid \setminus \mathrm{dx}$

$= {\int}_{1}^{2} {x}^{3} \ln x \setminus \mathrm{dx}$, as $x > 0$

$= {\int}_{1}^{2} \frac{d}{\mathrm{dx}} \left({x}^{4} / 4\right) \ln x \setminus \mathrm{dx}$

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

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

$= {\left[{x}^{4} / 4 \ln x - {x}^{4} / 16\right]}_{1}^{2}$

$= \left[\frac{16}{4} \ln 2 - \frac{16}{16}\right] - \left[\frac{1}{4} \left(0\right) - \frac{1}{16}\right]$

$= 4 \ln 2 - \frac{15}{16}$