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This question already has an answer here:

How to figure out the following integral? I have not been able to solve it from some time. $$\int_0^1\frac{x^{2} -1}{\log x}\,dx$$

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marked as duplicate by heropup, user7530, Pedro Tamaroff, Cookie, Hakim Jun 18 '14 at 18:20

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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}

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    $\begingroup$ @user152749 See here for more information on Feynman's method. (This problem happens to be worked out in this paper). $\endgroup$ – Alex Schiff Jun 18 '14 at 18:14
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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.

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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.

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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.)

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