TITLE: Evaluate $\int_0^1\frac{x^3 - x^2}{\ln x}\,\mathrm dx$?
QUESTION [9 upvotes]: How do I evaluate the following integral?
$$\int_0^1\frac{x^3 - x^2}{\ln x }\,\mathrm dx$$

REPLY [3 votes]: 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}