Evaluate $\int_0^1\frac{x^3 - x^2}{\ln x}\,\mathrm dx$? [duplicate]

How do I evaluate the following integral?

$$\int_0^1\frac{x^3 - x^2}{\ln x }\,\mathrm dx$$

Sub $x=e^{-u}$, $dx = -e^{-u} du$. Then the integral is

$$\int_0^1 dx \frac{x^3-x^2}{\log{x}} = \int_0^{\infty} du \, \frac{e^{-3 u} - e^{-4 u}}{u} = \int_0^{\infty} du \, \int_3^4 dt \, e^{-u t} \\ = \int_3^4 dt \,\int_0^{\infty} du \, e^{-u t} = \int_3^4 \frac{dt}{t} = \log{\frac{4}{3}}$$

The change in the order of integration is justified by Fubini's Theorem.

$$ \begin{align} \int_0^1\frac{x^3-x^2}{\log(x)}\mathrm{d}x &=\int_0^1\frac{x^4-x^3}{\log(x)}\frac{\mathrm{d}x}{x}\\ &=\lim_{a\to0}\int_a^1\frac{x^4-x^3}{\log(x)}\frac{\mathrm{d}x}{x}\\ &=\lim_{a\to0}\left(\int_a^1\frac{x^4-1}{\log(x)}\frac{\mathrm{d}x}{x} -\int_a^1\frac{x^3-1}{\log(x)}\frac{\mathrm{d}x}{x}\right)\\ &=\lim_{a\to0}\left(\int_{a^4}^1\frac{x-1}{\log(x)}\frac{\mathrm{d}x}{x} -\int_{a^3}^1\frac{x-1}{\log(x)}\frac{\mathrm{d}x}{x}\right)\\ &=\lim_{a\to0}\int_{a^4}^{a^3}\frac{x-1}{\log(x)}\frac{\mathrm{d}x}{x}\\ &=\lim_{a\to0}\int_{a^4}^{a^3}\frac1{\log(x)}\mathrm{d}x -\lim_{a\to0}\int_{a^4}^{a^3}\frac1{\log(x)}\frac{\mathrm{d}x}{x}\\ &=0-\Big[\log(\log(x))\Big]_{a^4}^{a^3}\\[6pt] &=\log(4)-\log(3)\\[12pt] &=\log(4/3) \end{align} $$

$$\begin{align} \int_0^1\frac{x^3-x^2}{\ln x }\mathrm{d}x &=\int_0^1\frac{x^3-1-x^2+1}{\ln x }\mathrm{d}x\tag{1}\\ &=\int_0^1\frac{x^\color{blue}{3}-1}{\ln x }\mathrm{d}x-\int_0^1\frac{x^\color{red}{2}-1}{\ln x}\mathrm{d}x\tag{2}\\ &=\ln(\color{blue}{3}+1)-\ln(\color{red}{2}+1)\tag{3}\\ &=\ln\frac43\tag{4}\\ \end{align}$$

$$\large\int_0^1\frac{x^3-x^2}{\log(x)}\mathrm{d}x=\ln\frac43$$

$\text{Explanation : } (3 )$

Consider parametric integral $\displaystyle\quad\quad\quad I(\alpha)=\int_0^1{\frac{x^{\alpha}-1}{\ln x}\mathrm dx}$

We have $I(0)=0$, By differentiating w.r.t $\alpha$ we get

$$ I'(\alpha) =\int_0^1{\frac{x^{\alpha}\ln x}{\ln x}\,\mathrm dx} =\int_0^1{x^{\alpha}\,\mathrm dx} =\frac{x^{\alpha+1}}{\alpha+1} =\frac{1}{\alpha+1}$$ Integrating w.r.t. $\alpha$ and using $I(0)=0$ we get $$I(\alpha)=\int_0^1{\frac{x^{\alpha}-1}{\ln x}\,\mathrm dx}=\ln(\alpha+1)$$