Logarithm powers! (13)

$\large \displaystyle \int_0^1 x^2 \, \left(\ln \left(\frac{1}{x} \right) \right)^3 \, dx$

$\large \displaystyle = - \int_0^1 x^2 \left(\ln (x) \right) \, dx$

$\large \displaystyle \color{#EC7300}{\text{Let } x = e^{-y}} \implies \color{royalblue}{dx = - e^{-y} \, dy}$

$\large \displaystyle = \int_0^{\infty} e^{-3y} \, y^3 \, dy$

$\large \displaystyle \color{#69047E}{\text{Again Let } 3y = u} \implies \color{#BA33D6}{3 \, dy = du}$

$\large \displaystyle = \int_0^{\infty} \frac{u^3}{3^3} \times \frac{e^{-u}}{3} \, du$

$\large \displaystyle = \color{#D61F06}{\frac{1}{3^4}} \, \color{#3D99F6}{\Gamma (4)} \approx \color{#20A900}{\boxed{0.074074}}$

$\displaystyle J(a)=\int_{0}^{1}x^a dx = \frac{1}{a+1}$

$\displaystyle -J'''(2) = -\int_{0}^{1} x^2 \ln^3 x dx = -\frac{6}{81}$ , So $\displaystyle \int_{0}^{1} x^2 \ln^3(\frac{1}{x}) dx = \frac{6}{81}\approx 0.074$