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I have been given the following improper integral to evaluate

$$ \int_{0}^{1} \frac{x^2-1}{\log{x}} dx $$

I tried substituting $ u = \log{x} $ which gives the following improper integral $\displaystyle \int_{-\infty}^{0} \frac{e^{3u}-e^{u}}{u} du $ This integral can be evaluated by software like Wolfram but I am interested in the method. How does one solve the integral analytically? I have no real idea on this. Thanks to all helpers.

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$$\int_{-\infty}^{0} \frac{e^{3u}-e^{u}}{u} du=\int_{-\infty}^{0} \int_1^3 e^{xu}\ dx\ du=\int_1^3 \int_{-\infty}^{0} e^{xu}\ du\ dx=\int_1^3 \dfrac1x\ dx=\ln3$$

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