Definite Integral: $\int_0^1\frac{x^{2} -1}{\log x}\,dx$ How to figure out the following integral? I have not been able to solve it from some time.
$$\int_0^1\frac{x^{2} -1}{\log x}\,dx$$
 A: The change of variables $x=e^{-t}$ transforms the integral to
$$
\int_0^\infty \frac{e^{-t}-e^{-3t}}{t}dt=\ln3
$$ 
Because this is a particular case of Frullanis integrals.
A: Let us generalize the problem. We will evaluate
$$
\mathcal{I}(\alpha)=\int_0^1\frac{x^\alpha-1}{\ln x}\ dx\qquad;\qquad \alpha>-1.\tag1
$$
Now we apply Feynman's method (differentiate under the integral sign). Diferentiating both sides of $(1)$ yields
\begin{align}
\frac{\partial\mathcal{I}}{\partial\alpha}&=\int_0^1\frac{\partial}{\partial\alpha}\left[\frac{x^\alpha-1}{\ln x}\right]\ dx\\
\mathcal{I}'(\alpha)&=\int_0^1 x^\alpha\ dx\\
&=\left.\frac{x^{\alpha+1}}{\alpha+1}\right|_{x=0}^1\\
&=\frac{1}{\alpha+1}.\tag2
\end{align}
Integrating $(2)$ yields
\begin{align}
\mathcal{I}(\alpha)&=\int\frac{1}{\alpha+1}\ d\alpha\\
&=\ln(\alpha+1)+C.\tag3
\end{align}
In order to find out our constant of integration, we let $\alpha = 0$ so that our integrand is $0$, implying that $C = 0$. Letting $\alpha = 2$ will of course solve our original problem:
\begin{align}
\color{purple}{\int_0^1\frac{x^\alpha-1}{\ln x}\ dx}&\color{purple}{=\ln(\alpha+1)}\\
\int_0^1\frac{x^2-1}{\ln x}\ dx&=\large\color{blue}{\ln3}.
\end{align}
A: Replace the exponent of 2 by a free parameter p. Then if you differentiate w.r.t. p a factor of log(x) will come down canceling the log(x) in the denominator. Calculate that (primary school level) integral and then integrate both sides of the result w.r.t. p from 0 to 2 to obtain the original integral.
A: Write $x=e^{-u}$, so that the integral becomes
$$\int_0^\infty{e^{-u}-e^{-3u}\over u}du$$
Now think of this as the limit as $N\to\infty$ of
$$\int_0^N{1-e^{3u}\over u}du - \int_0^N{1-e^{-u}\over u}du$$
The substitution $v=3u$ in the first integral turns this into
$$\int_N^{3N}{1-e^{-u}\over u}du=\ln3-\int_N^{3N}{e^{-u}\over u}du$$
That final integral clearly tends to $0$ as $N\to\infty$.
(Remark:  When the upper limit $N$ is first introduced, neither integral separately has a limit as $N\to\infty$.  Only their difference has a limit.)
