I want solve $$\int \frac{dx}{(x^2-x)^x}$$.
thanks for help
2026-03-29 22:13:26.1774822406
How solve $\int \frac{dx}{(x^2-x)^x}$
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Hint:
$\int\dfrac{dx}{(x^2-x)^x}$
$=\int(x^2-x)^{-x}~dx$
$=\int(e^{\ln(x^2-x)})^{-x}~dx$
$=\int e^{-x\ln(x^2-x)}~dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^n(\ln(x^2-x))^n}{n!}dx$