Evaluate $\int_0^1\frac{x^3 - x^2}{\ln x}\,\mathrm dx$? How do I evaluate the following integral?
$$\int_0^1\frac{x^3 - x^2}{\ln x }\,\mathrm dx$$
 A: $$\begin{align}
\int_0^1\frac{x^3-x^2}{\ln x }\mathrm{d}x
&=\int_0^1\frac{x^3-1-x^2+1}{\ln x }\mathrm{d}x\tag{1}\\
&=\int_0^1\frac{x^\color{blue}{3}-1}{\ln x }\mathrm{d}x-\int_0^1\frac{x^\color{red}{2}-1}{\ln x}\mathrm{d}x\tag{2}\\
&=\ln(\color{blue}{3}+1)-\ln(\color{red}{2}+1)\tag{3}\\
&=\ln\frac43\tag{4}\\
\end{align}$$

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


$\text{Explanation : } (3
)$
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
$$I(\alpha)=\int_0^1{\frac{x^{\alpha}-1}{\ln x}\,\mathrm dx}=\ln(\alpha+1)$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large%
\int_{0}^{1}{x^{3} - x^{2} \over \ln\pars{x}}\,\dd x}
=\int_{0}^{1}\pars{x^{2} - x^{3}}\
\overbrace{\int_{0}^{\infty}x^{t}\,\dd t}^{\dsc{-\,{1 \over \ln\pars{x}}}}\
\,\dd x\ =\
\int_{0}^{\infty}\int_{0}^{1}\pars{x^{2 + t} - x^{3 + t}}\,\dd x\,\dd t
\\[5mm]&=\int_{0}^{\infty}\pars{{1 \over t + 3} - {1 \over t + 4}}\,\dd t
=\left.\ln\pars{t + 3 \over t + 4}\right\vert_{\, t\ =\ 0}^{\, t\ \to\ \infty}
=-\ln\pars{3 \over 4}=\color{#66f}{\large\ln\pars{4 \over 3}}
\end{align}
A: You may like this method. Note
$$ \lim_{n\to\infty} n(x^{\frac{1}{n}}-1)=\ln x, \text{ for }x>0 $$
and hence
\begin{eqnarray}
\int_0^1\frac{x^3-x^2}{\ln x}dx&=&\int_0^1\lim_{n\to\infty}\frac{x^3-x^2}{n(x^{\frac{1}{n}}-1)}dx\\
&=&\lim_{n\to\infty}\int_0^1\frac{x^2(x-1)}{n(x^{\frac{1}{n}}-1)}dx\\
&=&\lim_{n\to\infty}\int_0^1x^2\frac{1}{n}\sum_{k=0}^{n-1}x^{\frac{k}{n}}dx\\
&=&\lim_{n\to\infty}\int_0^1\frac{1}{n}\sum_{k=0}^{n-1}x^{\frac{k}{n}+2}dx\\
&=&\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{\frac{k}{n}+3}\\
&=&\int_0^1\frac{1}{x+3}dx\\
&=&\ln\frac{4}{3}.
\end{eqnarray}
A: 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.
A: $$
\begin{align}
\int_0^1\frac{x^3-x^2}{\log(x)}\mathrm{d}x
&=\int_0^1\frac{x^4-x^3}{\log(x)}\frac{\mathrm{d}x}{x}\\
&=\lim_{a\to0}\int_a^1\frac{x^4-x^3}{\log(x)}\frac{\mathrm{d}x}{x}\\
&=\lim_{a\to0}\left(\int_a^1\frac{x^4-1}{\log(x)}\frac{\mathrm{d}x}{x}
-\int_a^1\frac{x^3-1}{\log(x)}\frac{\mathrm{d}x}{x}\right)\\
&=\lim_{a\to0}\left(\int_{a^4}^1\frac{x-1}{\log(x)}\frac{\mathrm{d}x}{x}
-\int_{a^3}^1\frac{x-1}{\log(x)}\frac{\mathrm{d}x}{x}\right)\\
&=\lim_{a\to0}\int_{a^4}^{a^3}\frac{x-1}{\log(x)}\frac{\mathrm{d}x}{x}\\
&=\lim_{a\to0}\int_{a^4}^{a^3}\frac1{\log(x)}\mathrm{d}x
-\lim_{a\to0}\int_{a^4}^{a^3}\frac1{\log(x)}\frac{\mathrm{d}x}{x}\\
&=0-\Big[\log(\log(x))\Big]_{a^4}^{a^3}\\[6pt]
&=\log(4)-\log(3)\\[12pt]
&=\log(4/3)
\end{align}
$$
A: We have:
$$I = \int_{0}^{1} \frac{x^2(x - 1)}{\log(x)}$$
Consider:
$$I(b) = \int_{0}^{1} \frac{x^b(x - 1)}{\log(x)}$$
Recognize that $I(0) = \int_{0}^{1} (x-1)/\log(x) dx = \log(2)$, this is a particular solution for later.
Differentiate partially with respect to $b$
$$I'(b) = \int_{0}^{1} \frac{\log(x)\cdot x^b(x - 1)}{\log(x)} dx$$
$$I'(b) = \int_{0}^{1} \frac{x^b(x - 1)}{1} dx = \int_{0}^{1} x^b(x - 1) dx $$
$$I'(b) = \int_{0}^{1} x^{b+1} - x^b dx$$
$$I'(b) = \frac{1}{b+2} - \frac{1}{b+1}$$
Integrate this with respect to $db$
$$I(b) = \int \frac{1}{b+2} - \frac{1}{b+1} db$$
$$I(b) = \log(b+2) - \log(b+1) + C$$
$C= 0$
$$I(2) = \log(4) -  \log(3) = \log(4/3)$$
