### Video Transcript

Determine the indefinite integral
of the fifth root of π¦ squared divided by the cube root of π¦ with respect to
π¦.

In this question, weβre asked to
evaluate the integral of the quotient of two functions. And our integrand is given in terms
of π¦, and weβre integrating with respect to π¦. Remember, as long as our integrand
and what weβre integrating with respect to share the same variable, it doesnβt
matter what we call this variable. So, we can integrate this using all
of our standard rules. However, as it stands, we donβt
know how to integrate this function directly. So, weβre going to need to rewrite
this in a form which we can integrate. And since this expression involves
a lot of exponents, weβll do this by using our laws of exponents.

The first one of our laws of
exponents weβll use is the following. We know the πth root of π₯ is
equal to π₯ to the power of one over π. Of course, this is only true for
positive integers π. And in fact, we can use this to
rewrite both the numerator of our integrand and the denominator. First, in our numerator, the value
of π is five. So, we get π¦ squared all raised to
the power of one over five. Then, in our denominator, we can
rewrite the cube root of π¦ as π¦ to the power of one over three.

So, weβve rewritten our integral in
the following form. However, we still canβt evaluate
this integral. So, weβre going to need to simplify
this even further. We can see in our numerator we have
π¦ squared all raised to the power of one over five. And we can rewrite this as π¦
raised to the power of some number by using our laws of exponents. Recall, π raised to the power of
π all raised to the power of π is equal to π raised to the power of π times
π. Weβll apply this to rewrite the
numerator of our integrand. Our value of π is π¦, our value of
π is two, and our value of π is one over five. This gives us a new numerator of π¦
to the power of two times one over five.

So, now, we have the integral of π¦
to the power of two times one over five divided by one to the power of one-third
with respect to π¦. And of course, we can simplify the
exponents in our numerator. Two times one over five is equal to
two over five. Now, we can see our integrand is
the quotient of two exponents where π¦ is the base in both of these. And once again, we can simplify
this by using our laws of exponents. We know π¦ raised to the power of
π divided by π¦ raised to the power of π is equal to π¦ raised to the power of π
minus π.

In this case, our value of π is
two over five and our value of π is one-third. So using this, we get the integral
of π¦ over two over five minus one-third with respect to π¦. And we can simplify this. Evaluating our exponent, we get two
over five minus one-third is equal to one over 15. So, after all of this manipulation,
we were able to rewrite our integral as the integral of π¦ raised to the power of
one over 15 with respect to π¦. And now, we can evaluate this
integral by using the power rule for integration. Recall for all real constants π
and π, where π is not equal to negative one, the integral of π times π¦ to the
πth power with respect to π¦ is equal to π times π¦ to the power of π plus one
divided by π plus one plus our constant of integration πΆ.

So, we want to add one to our
exponent of π¦ and then divide by this new exponent. We can see in our case the exponent
of π¦ is one over 15. So, we need to add one to this
exponent of one over 15 and then divide by this new exponent. This gives us π¦ to the power of
one over 15 plus one all divided by one over 15 plus one. And of course, we need to add our
constant of integration πΆ.

And the last thing weβll do is
simplify our exponent and our denominator. One over 15 plus one is equal to 16
over 15. This gives us π¦ to the power of 16
over 15 divided by 16 over 15 plus πΆ. And we can do one last piece of
simplification. Instead of dividing by 16 over 15,
weβll multiply by the reciprocal. And doing this gives us our final
answer.

Therefore, we were able to show the
integral of the fifth root of π¦ squared over the cube root of π¦ with respect to π¦
is equal to 15 times π¦ to the power of 16 over 15 divided by 16 plus πΆ.