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$$\int_{0}^{1} \frac{(x^{2}-1)}{\ln x} \ dx$$ I tried 3 different methods: First , I tried to use the substitution $x = 1/t$ which led to integral of $e^t /t$ form which I was unable to integrate analytically Then I tried to use integration by parts taking ln as the second function and $x^2-1$ as the second function but it did not help Is there any way to evaluate it without using the residue theorem and differentiation under integral sign??

conan
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The Grim Reaper
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1 Answers1

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$$\int_0^1\frac{x^2-1}{\log x}dx = \int_0^1\int_0^2x^y\:dydx = \int_0^2\int_0^1x^ydxdy = \int_0^2\frac{dy}{y+1} = \boxed{\log 3}$$

Ninad Munshi
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