Using the method of $u-$substitution,

${\displaystyle \underset{1}{\overset{5}{}}}(3x-4{)}^{5}dx={\displaystyle \underset{a}{\overset{b}{}}}f(u)du$

where

$u=(enter$ a function $ofx)$

$du=,dx(enter$ a function $ofx)$

$a=(enter$ a number$)$

$b=\frac{u^5}{5}(enter$ a number$)$

$f(u)=(enter$ a function $ofu).$

The value of the original integral is $$

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