How do I evaluate the following integral?  $$\int_0^1\frac{x^3 - x^2}{\ln x }\,\mathrm dx$$
I would like to show that   $$ \int_{0}^{1} \frac{x-1}{\ln(x)} \mathrm dx=\ln2 $$  What annoys me is that $ x-1 $ is the numerator so the geometric power series is useless.  Any idea?
I stumbled upon  on last weekend, it introduces an integral trick that exploits . On its last page, the author, Mr. Anonymous, left several exercises without any hints, one of them is to evaluate the Gaussian integral $$ \int^\infty_0 e^{-x^2} \,dx= \frac{\sqrt{\pi}}{2} $$ using this parametrization trick. I had been evaluating it through trial and error using different paramatrizations, but no luck so far.     Here are what I have tried so far:   A first instinct would be do something like:$$ I(b) = \int^\infty_0 e^{-f(b)x^2}\,dx $$ for some permissible function $f(\cdot)$, differentiating it will lead to a simple solvable ode: $$ \frac{I'(b)}{I(b)} = -\frac{f'(b)}{2f(b)} $$ which gives: $$ I(b) = \frac{C}{\sqrt{f(b)}}. $$ However, finding this constant $C$ basically is equivalent to evaluating the original integral, we are stuck here without leaving this parametrization trick framework. A second try involves an exercise on the same page: $$ I(b) = \int^\infty_0 e^{-\frac{b^2}{x^2}-x^2}dx. $$ Taking derivative and rescaling the integral using change of variable we have: $$ I'(b) = -2I(b). $$ This gives us another impossible to solve constant $C$ in: $$ I(b) = C e^{-2b} $$ without leaving this framework yet again. The third try is trying modify Américo Tavares's answer in this : $$ I(b) = \int^\infty_0 be^{-b^2x^2}\,dx. $$ It is easy to show that: $$ I'(b) = \int^\infty_0 e^{-b^2x^2}\,dx - \int^\infty_0 2b^2 x^2 e^{-b^2x^2}\,dx = 0 $$ by an integration by parts identity: $$ \int^\infty_0 x^2 e^{- c x^2}\,dx = \frac{1}{2c}\int^\infty_0 e^{- c x^2}\,dx . $$ Then $I(b) = C$, ouch, stuck again at this constant.     Notice in that  question, Bryan Yocks's answer is somewhat similar to the idea of parametrization, however he has to introduce another parametric integration to produce a definite integral leading to $\arctan$.  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