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Q. $\int\limits_{0}^{1} \frac{x d x}{(1-x)^{3 / 4}}=$

Integrals

Solution:

$\int\limits_{0}^{1} \frac{x d x}{(1-x)^{3 / 4}}=-\int\limits_{0}^{1} \frac{(1-x)-1}{(1-x)^{3 / 4}} d x$
$=-\int\limits_{0}^{1}\left[(1-x)^{1 / 4}-(1-x)^{-3 / 4}\right] d x$
$=\left|\frac{(1-x)^{1 / 4}}{\frac{1}{4}}-\frac{(1-x)^{5 / 4}}{\frac{5}{4}}\right|_{0}^{1}$
$=-4(0-1)+\frac{4}{5}(0-1)$
$=-\frac{16}{5}$