# Question #092c3

##### 1 Answer
Nov 5, 2017

${\int}_{0}^{1} {x}^{3} \cdot {\left[L n \left(\frac{1}{x}\right)\right]}^{3} \cdot \mathrm{dx}$=$\frac{3}{128}$

#### Explanation:

${\int}_{0}^{1} {x}^{3} \cdot {\left[L n \left(\frac{1}{x}\right)\right]}^{3} \cdot \mathrm{dx}$

After using $u = L n \left(\frac{1}{x}\right) = - L n x$, $L n x = - u$, $x = {e}^{- u}$ and $\mathrm{dx} = - {e}^{- u} \cdot \mathrm{du}$ transforms, this integral became,

${\int}_{\infty}^{0} - {u}^{3} \cdot {e}^{- 4 u} \cdot \mathrm{du}$

$= {\int}_{0}^{\infty} {u}^{3} \cdot {e}^{- 4 u} \cdot \mathrm{du}$

After using tabular integration,

$= {\int}_{0}^{\infty} {u}^{3} \cdot {e}^{- 4 u} \cdot \mathrm{du}$

=${\left[{u}^{3} \cdot \left(- \frac{1}{4} {e}^{- 4 u}\right) - 3 {u}^{2} \cdot \frac{1}{16} {e}^{- 4 u} + 6 u \cdot \left(- \frac{1}{64} \cdot {e}^{- 4 u}\right) - 6 \cdot \frac{1}{256} {e}^{- 4 u}\right]}_{0}^{\infty}$

=${\left[- \frac{1}{128} {e}^{- 4 u} \cdot \left(32 {u}^{3} + 24 {u}^{2} + 12 u + 3\right)\right]}_{0}^{\infty}$

=$\frac{3}{128}$