Question Video: Finding the Integration of a Function Involving Exponential Functions by Distributing the Division | Nagwa Question Video: Finding the Integration of a Function Involving Exponential Functions by Distributing the Division | Nagwa

# Question Video: Finding the Integration of a Function Involving Exponential Functions by Distributing the Division Mathematics • Third Year of Secondary School

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Determine β« (8π^(2π₯) π₯ + π^(π₯))/(5π^(π₯) π₯) dπ₯.

03:16

### Video Transcript

Determine the integral of eight π to the power of two π₯ times π₯ plus π to the power of π₯ all divided by five π to the power of π₯ times π₯ with respect to π₯.

Weβre asked to evaluate the integral of the quotient of two functions. And we donβt know how to do this in the general case. So weβre going to need to come up with a method of integrating this. One thing we should always check is, can we simplify our integrand? And one thing we can notice about our integrand is our denominator only contains one term. This means we could try dividing each term in our numerator separately by our denominator. So weβll start by splitting our integrand into two. And now we can see we can simplify each of the terms inside of our integrand separately.

In the first term, we have a shared factor of π to the power of π₯ in our numerator and our denominator and a shared factor of π₯ in our numerator and our denominator. We can cancel all of these to get eight π to the power of π₯ all divided by five. And one thing thatβs worth reiterating here is, why weβre allowed to cancel the shared factor of π₯ divided by π₯? Weβre calculating an indefinite integral. This means weβre looking for an antiderivative of our function. And this antiderivative will only be valid for certain values of π₯. If we have π₯ divided by π₯ in our integrand, then we know that this wonβt be valid when π₯ is equal to zero. So weβre allowed to cancel our shared factors of π₯. And we just need to be careful because we know this will change the values of π₯ for which our answer is valid.

Letβs now move on to the second term in our integrand. Once again, we can see thereβs a shared factor of π to the power of π₯ in our numerator and our denominator. So weβll cancel these out. So by splitting our integrand into two and canceling the shared factors, we were able to rewrite our integral as the integral of eight π to the power of π₯ over five plus one over five π₯ with respect to π₯. And at this point, we can integrate this term by term. Letβs start with the integral of our first term. We can do this by using our rules for integrating exponential functions.

We know for any real constant π, the integral of ππ to the power of π₯ with respect to π₯ is equal to ππ to the power of π₯ plus a constant of integration πΆ. In our case, the value of π is equal to eight over five. It might be easier to see this by rewriting this term as eight over five times π to the power of π₯. So, by setting π equal to eight over five, we can integrate our first term to get eight π to the power of π₯ divided by five. And itβs worth pointing out we donβt need to add a constant of integration for each term. We can combine all of these into one.

Now we just need to evaluate the integral of our second term. And weβll do this by recalling one of our rules for integrating reciprocal functions. We know for any real constant π, the integral of π over π₯ with respect to π₯ is equal to π times the natural logarithm of the absolute value of π₯ plus a constant of integration πΆ. And we can evaluate the integral of this term by setting π equal to one over five. It might be easier to see this to write one over five π₯ as one-fifth times one over π₯.

So by setting the value of π equal to one-fifth, we can integrate our second term to get one-fifth times the natural logarithm of the absolute value of π₯. And remember, we need to add a constant of integration weβll call πΆ. And this gives us our final answer. Therefore, we were able to show the integral of eight π to the power of two π₯ times π₯ plus π to the power of π₯ all divided by five π to the power of π₯ times π₯ with respect to π₯ is equal to eight π to the power of π₯ over five plus one-fifth times the natural logarithm of the absolute value of π₯ plus πΆ.

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