### Video Transcript

Determine the integral of three π₯
plus four all squared multiplied by π to the power of π₯ with respect to π₯.

We need to determine the value of
an indefinite integral. We can see that our integrand is
the product of two functions. Weβve seen a few different ways of
dealing with integrals of this form. For example, we might be tempted to
try integration by substitution. However, thereβs no obvious
substitution that would make this easier. Instead, we could try using
integration by parts.

First, we recall integration by
parts tells us the integral of π’π£ prime with respect to π₯ is equal to π’ times π£
minus the integral of π£ times π’ prime with respect to π₯. So, integration by parts gives us a
method of integrating the product of π’ and π£ prime with respect to π₯.

But, remember, we need to choose
our function π’. Thereβs a few different ways of
choosing our function π’. Usually, we pick the function which
differentiates to give a simpler function. Using this logic, we see the
derivative of π to the power of π₯ is just equal to π to the power of π₯. This doesnβt get any simpler. However, we know three π₯ plus four
all squared is a quadratic. So, when we differentiate this with
respect to π₯, weβll get a linear function. And a linear function is more
simple than a quadratic. So, this makes a good candidate for
π’.

Alternatively, we couldβve used the
LIATE method. Since there are no logarithmic or
inverse trigonometric functions in our integrand, we would choose the algebraic
function three π₯ plus four all squared to be π’. Both of these methods pick the same
function. So, weβll set π’ to be three π₯
plus four all squared and π£ prime to be π to the power of π₯.

To use integration by parts, we
also need expressions for π’ prime and π£ prime. Letβs start with finding π’
prime. To find π’ prime, we need to
differentiate three π₯ plus four all squared. We could do this by using the chain
rule or the general power rule. However, because weβre only
squaring this linear function, we could do this by using the FOIL method or binomial
expansion. We would get that π’ is equal to
nine π₯ squared plus 24π₯ plus 16. And now, we can differentiate this
term by term by using the power rule for differentiation.

We need to multiply it by our
exponents of π₯ and then reduce this exponent by one. This gives us that π’ prime is
equal to 18π₯ plus 24. Now, we need to find our function
π£. π£ will just be an antiderivative
of π to the power of π₯. Well, we know the derivative of π
to the power of π₯ with respect to π₯ is just equal to π to the power of π₯. So, we can just set π£ to be π to
the power of π₯.

Now that we found expressions for
π’ prime and π£, we can use integration by parts to attempt to evaluate our
integral. Substituting in our expressions for
π’, π£, π’ prime, and π£ prime, we get that our integral is equal to nine π₯ squared
plus 24π₯ plus 16 all multiplied by π to the power of π₯ minus the integral of π
to the π₯ times 18π₯ plus 24 with respect to π₯.

But now we can see a problem. We have an integral in this
expression which we still canβt integrate directly. But now we can see something
interesting. Our original integrand was a
quadratic multiplied by π to the power of π₯. However, we now have just a linear
function multiplied by π to the power of π₯. If we did this process one more
time, then our linear function would just become a constant. Then, we could just evaluate our
integral.

So, weβre gonna use integration by
parts on the integral of π to the power of π₯ times 18π₯ plus 24 with respect to
π₯. This time, our function π’ will be
18π₯ plus 24, and π£ prime will be equal to π to the power of π₯. And just as we did before, we can
find expressions for π’ prime and π£ prime. π’ prime will be the derivative of
18π₯ plus 24 with respect to π₯, which we know is 18. And π£ will be π to the power of
π₯.

Now, all we need to do is
substitute our expressions for π’, π£, π’ prime, and π£ prime into our formula for
integration by parts. Doing this, we get 18π₯ plus 24 all
multiplied by π to the power of π₯ minus the integral of π to the power of π₯
times 18 with respect to π₯. And we can rearrange our integrands
to give us 18π to the power of π₯. And now, we can just evaluate this
integral.

Negative one times the integral of
18π to the power of π₯ with respect to π₯ is equal to negative 18 times π to the
power of π₯ plus a constant of integration πΆ. So, this gave us 18π₯ plus 24 all
multiplied by π to the power of π₯ minus 18π to the power of π₯ plus πΆ. And before we substitute this
expression back into our original integral, we can take out a factor of π to the
power of π₯. Taking out this common factor of π
to the power of π₯, we get 18π₯ plus 24 minus 18 all multiplied by π to the power
of π₯ plus πΆ.

And of course, we can now simplify
24 minus 18 to give us six. So, we simplified this expression
to give us 18π₯ plus six all multiplied by π to the power of π₯ plus πΆ. We can now substitute this
expression back into our original integral. Doing this, we get nine π₯ squared
plus 24π₯ plus 16 all multiplied by π to the power of π₯ minus 18π₯ plus six times
π to the power of π₯ plus our constant of integration πΆ.

We now want to distribute this
negative one over our parentheses. But we see this will give us the
term negative πΆ. We could call this negative πΆ. However, πΆ is our constant of
integration. We can call this whatever we
want. So, we couldβve originally called
this negative πΆ, so we can just write negative πΆ as πΆ. This gives us nine π₯ squared plus
24π₯ plus 16 all multiplied by π to the power of π₯ minus 18π₯ plus six times π to
the power of π₯ plus our constant of integration πΆ.

And we could leave our answer like
this. However, we see that our first two
terms both share a factor of π to the power of π₯. So, by taking out this common
factor of π to the power of π₯, we get nine π₯ squared plus 24π₯ plus 16 minus 18π₯
minus six all multiplied by π to the power of π₯ plus πΆ. And now, we can simplify this
expression. We have 24π₯ minus 18π₯ is equal to
six π₯. And 16 minus six is just equal to
10.

And then, by rewriting π to the
power of π₯ at the start of our first factor, we got π to the power of π₯ times
nine π₯ squared plus six π₯ plus 10 plus the constant of integration πΆ. And this is our final answer. Therefore, by using integration by
parts twice, we were able to determine that the integral of three π₯ plus four all
squared times π to the power of π₯ with respect to π₯ is equal to π to the power
of π₯ times nine π₯ squared plus six π₯ plus 10 plus the constant of integration
πΆ.