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Integrate $f(x)$ from 0 to 1 where $f(x) = \frac{x^3-1}{lnx}$

I received this problem and a variety of others in an advanced mathematics exam. I tried a classical trigonometric substituition approach with $x=sec^\frac{2}{3}t$ but it gets really long and still gives no answer. I tried multiplying the numerator and denominator by $x$ and trying $ln(lnx)=z$ substitution as $dz=\frac{1}{lnx.x}dx$. I am not able to see which definite integral properties would actually help me here, either. Please advise.

Sat D
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    $$\int_0^1 \frac{x^3 - 1}{\ln x},dx = \int_0^1 \int_0^{3} e^{y\ln x},dy,dx = \int_0^3 \int_0^1 x^y,dx,dy = \int_0^3 \frac{1}{1+y},dy$$ – r9m Dec 02 '15 at 10:37
  • @r9m. Nice solution, indeed ! I was, just being stupid, thinking about the antiderivative. – Claude Leibovici Dec 02 '15 at 10:39
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    See also here http://math.stackexchange.com/questions/1082626/how-to-find-the-value-of-int-01-fracx-1-log-x-dx/1082642#1082642 – Kelenner Dec 02 '15 at 10:39
  • Ohh right. The parameter trick. Damn. Thanks for reminding me! – Sat D Dec 02 '15 at 10:42

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