Question

$\displaystyle \sf \red{\int_{0}^{1} (logx) {}^{2020} \: dx}$​

Answer 1

Answer:

$\huge \green{2021!}$

Step-by-step explanation:

Put

$log \: x \: = - t \\ \\ \frac{1}{x} dx = - dt \implies \: dx = - {e}^{ - t} dt$

Integral becomes

$\int_{ \infty }^{0} - ( - t) {}^{2020} {e}^{ -t} dt \\ \\ = \int_{ 0}^{ \infty } t {}^{2020} {e}^{ -t} \: dt \\ \\ = \Gamma(2021)$

Answer 2

$\huge\mathfrak\red{Thank \: You}$