# Video: Finding the Integration of an Exponential Function Multiplied by a Polynomial Using Integration by Parts

Determine β«(3π₯ + 4)Β²π^π₯dπ₯.

06:32

### 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. Weβve seen a few different ways of dealing with integrals of this form. For example, we might be tempted to try integration by substitution. However, thereβs 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’. Thereβs 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 doesnβt get any simpler. However, we know three π₯ plus four all squared is a quadratic. So, when we differentiate this with respect to π₯, weβll get a linear function. And a linear function is more simple than a quadratic. So, this makes a good candidate for π’.

Alternatively, we couldβve 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, weβll 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. Letβs 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 weβre 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 canβt 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, weβre 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 πΆ. So, this gave us 18π₯ plus 24 all multiplied by π to the power of π₯ minus 18π to the power of π₯ plus πΆ. 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 πΆ.

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 πΆ. 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 πΆ.

We now want to distribute this negative one over our parentheses. But we see this will give us the term negative πΆ. We could call this negative πΆ. However, πΆ is our constant of integration. We can call this whatever we want. So, we couldβve originally called this negative πΆ, so we can just write negative πΆ as πΆ. 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 πΆ.

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 πΆ. 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 πΆ. 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 πΆ.