### Video Transcript

Determine the indefinite
integral having negative four multiplied by the fifth root of π₯ to the power of
nine plus eight all multiplied by the fifth root of π₯ squared with respect to
π₯.

Letβs start by writing this
roots as powered. We know that the πth root of
π₯ is equal to π₯ to the power of one over π. Once weβve written our roots as
powers, we can then combine them with the existing powers, using the fact that
π₯ to the power of π to the power of π is equal to π₯ to the power of π
multiplied by π. Therefore, π₯ to the power of
nine to the power of one-fifth becomes π₯ to the power of nine-fifths. And π₯ squared to the power of
one-fifths becomes π₯ to the power of two-fifths. Now, we can expand the
brackets, using the fact that π₯ to the power of π times π₯ to the power of π
is equal to π₯ to the power of π plus π. So our integral becomes the
integral of negative four π₯ to the power of eleven-fifths plus eight multiplied
by π₯ to the power of two-fifths with respect to π₯.

Here, we can use the power rule
for integration which tells us that the indefinite integral of π₯ to the power
of π with respect to π₯ is equal to π₯ to the power of π plus one over π plus
one plus πΆ. We can apply this rule to our
integral term by term. For the first term, we have
negative four π₯ to the power of eleven-fifths. Therefore, π is equal to
eleven-fifths. When we integrate this term, we
get negative four multiplied by π₯ to the power of π plus one. And π plus one is simply
sixteen-fifths. So itβs π₯ to the power of
sixteen-fifths. And then we need to divide by
π plus one. So thatβs dividing by
sixteen-fifths.

For the second time, we have
eight multiplied by π₯ to the power of two-fifths. Therefore, π is
two-fifths. So we add eight multiplied by
π₯ to the power of π plus one which is π₯ to the power of seven-fifths. And we then divide by
seven-fifths. And we mustnβt forget to add
our constant of integration πΆ. Now, all that remains to do is
to simplify. And so we obtain a solution
that the indefinite integral of negative four multiplied by the fifth root of π₯
to the power of nine plus eight all multiplied by the fifth root of π₯ squared
with respect to π₯ is equal to negative five multiplied by π₯ to the power of
sixteen-fifths over four plus 40 multiplied by π₯ to the power of seven-fifths
over seven plus πΆ.