### Video Transcript

Determine the indefinite
integral of π₯ minus six multiplied by π₯ minus five multiplied by π₯ minus
three.

Letβs start by expanding the
brackets. Expanding the first two sets of
brackets, we get π₯ squared minus 11π₯ plus 30. And we then multiply this by π₯
minus three. We obtain the indefinite
integral of π₯ cubed minus 14π₯ squared plus 63π₯ minus 90 which is in fact a
polynomial. And we can use the power rule
for integration to integrate this term by term. The power rule tells us that
the 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. If instead we were integrating
π₯ to the power of π multiplied by some constant π, then since we can factor a
constant out of our integral, then this would simply be equal to π multiplied
by π₯ to the power of π plus one over π plus one plus πΆ.

Now, you may be wondering why
we havenβt multiplied the πΆ by π. And thatβs because π is also a
constant. Therefore, π multiplied by πΆ
is a constant. And we can just rename this new
constant to be πΆ. So now, letβs apply this rule
to our integral term by term. The first term is π₯ cubed. Therefore, π is equal to
three. We increase the power by one
and divide by the new power to get π₯ to the power of four over four. The next term is negative 14π₯
squared. Negative 14 is just a
constant. So that will remain. Our power is two. So π is two. We increase the power by one to
get π₯ cubed and divide by the new power.

On next term, we have 63π₯. So we can start by writing in
our constant of 63. We then note that π₯ is equal
to π₯ to the power of one. So π is equal to one. So when we integrate this, we
get π₯ squared over two. For our final term, we have
negative 90. And we know that this can also
be written as negative 90 multiplied by π₯ to the power of zero since π₯ to the
power of zero is just one. So now, we can integrate
it. We start by writing our
constant of negative 90. Since our power of π₯ is zero,
we increase the power by one giving us π₯ to the power of one and divide by the
new power. So thatβs dividing by one. And we miss then adding our
constant of integration πΆ.

We can write this out a little
neater for our solution which is π₯ to the power of four over four minus 14π₯
cubed over three plus 63π₯ squared over two minus 90π₯ plus πΆ.