Question Video: Finding the Integration of a Function Involving an Exponential Function Using Integration by Parts Mathematics

Video Transcript

Determine the integral of nine 𝑥 plus seven over 𝑒 to the power of five 𝑥 d𝑥.

So now the first thing we want to do before we integrate this is we’re actually gonna rewrite it. I’ve actually rewritten it as the integral of nine 𝑥 plus seven multiplied by 𝑒 to the power of negative five 𝑥 d𝑥. So what I’ve done is I’ve actually made 𝑒 to the power of negative five 𝑥 instead of one over 𝑒 to the power five 𝑥 because that’s actually the same thing.

Okay, great! So now what I’m gonna do is actually use integration by parts to actually help us to integrate this. And what integration by parts tells us is that if we actually have something in the form of the integral of 𝑢 d𝑣 d𝑥 d𝑥, and therefore it’s gonna be equal to 𝑢𝑣 minus the integral of the 𝑣 d𝑢 d𝑥 d𝑥. Okay, so now we’ve actually got this let’s use it to actually determine what the integral of our expression is going to be.

So first of all, we need to decide what are 𝑢 and what are d𝑣 d𝑥 are going to be. So our 𝑢 is going to be nine 𝑥 plus seven, and our d𝑣 d𝑥 is going to be equal to 𝑒 to the power of negative five 𝑥. So okay, great! So now what we want to do is actually to differentiate 𝑢 to find d𝑢 d𝑥, and then we want to integrate d𝑣 d𝑥 to actually find 𝑣. So therefore if we differentiate nine 𝑥 plus seven, we just get nine.

And that’s because nine 𝑥 differentiates to nine because actually it’s the coefficient, which is nine, multiplied by the exponent, which is one, which gives us nine. And then we reduce exponent of our 𝑥 so it goes to zero. So we’re just left with nine. And then seven if we differentiate that, it goes to zero. So d𝑢 d𝑥 equals nine and we can say that 𝑣 is equal to negative 𝑒 to the power of negative five 𝑥 over five. And we got that by integrating 𝑒 to the power of negative five 𝑥. That’s because we know that actually if you integrate 𝑒 to the power of 𝑘𝑥 then what you actually gonna get is one over 𝑘 𝑒 to the power of 𝑘𝑥. So we applied that and we got 𝑣 is equal to negative 𝑒 to the negative five 𝑥 over five.

Okay, great! So now we got 𝑢, 𝑣, d𝑢 d𝑥, and d𝑣 d𝑥. So now we can actually apply what we’ve seen in the integration by parts. So we can say that actually the integral that we’re looking for is equal to negative nine 𝑥 plus seven multiplied by 𝑒 to the power of negative five 𝑥 over five, because this is our 𝑢 multiplied by our 𝑣, and then minus the integral of negative nine 𝑒 to the power of negative five 𝑥 over five d𝑥. And this is because this is our 𝑣 d𝑢 d𝑥.

Okay, great! So now let’s integrate the second part. So therefore, we’re gonna have negative nine 𝑥 plus seven multiplied by 𝑒 to power of negative five 𝑥 over five, and then plus nine over 25 𝑒 to the power of negative five 𝑥. We’ve got that because we actually integrated our negative nine 𝑒 to the power of negative five 𝑥 over five using the same method we’d used previously.

So just a quick recap what we’ve done. So we integrate our negative nine 𝑒 to the power of negative five 𝑥 over five d𝑥. What we get is that this is equal to well we have one over our 𝑘 and our 𝑘 was our coefficient of d𝑥 in the exponent, so one over negative five, multiplied by negative nine 𝑒 to the power of negative five 𝑥 over five. And when we did that, we actually have the denominators multiplied to get us negative five multiplied by five, which gives us negative 25. Because it was a negative nine 𝑒 to the power of negative five 𝑥 over five becomes positive, so we get positive nine 𝑒 to the power of negative five 𝑥 over 25.

Okay, great! So we’ve now done all the integrating. What we need to do is just simplify. It’s worth noting also this point that what we will need is actually a constant of integration. So now what we want to do is actually complete the next final steps, which is actually to simplify what we’ve got. And what I’ve done here is I’ve included our constant of integrations, so I’ve got plus 𝐶. So the first thing we want to do is actually well we want to add our fractions. And to enable us to do that, what we need to do is actually have the same denominator.

So I’ve multiplied the first fraction numerator and denominator by five. So we have negative five multiplied by nine 𝑥 plus seven 𝑒 to the power of negative five 𝑥 plus nine 𝑒 to the power of negative five 𝑥 over 25. So therefore, if we actually expand the parentheses, what we get is negative 45𝑒 to the power of negative five 𝑥 𝑥 plus 35𝑒 to the power of negative five 𝑥 plus nine 𝑒 to the negative five 𝑥 over 25.

So therefore, if we actually collect terms on numerator, we get negative 45𝑒 to the power of negative five 𝑥 𝑥 plus 44𝑒 to the negative five 𝑥 over 25. And then what we can do is actually take out 𝑒 to negative five 𝑥 as a factor. So we get negative 45𝑥 plus 44 multiplied by 𝑒 to negative five 𝑥 over 25, so which leads us to our final answer where we’ve actually taken negative nine over five out as a factor. So we can say that the integral of nine 𝑥 plus seven over 𝑒 to the power of five 𝑥 d𝑥 is equal to negative nine over five multiplied by 𝑥 plus 44 over 45 multiplied by 𝑒 to the power of negative five 𝑥, then plus 𝐶 because we don’t forget our constant of integration. And that is our final answer.