Video Transcript
Determine the integral of two plus three π to the power of π₯ all divided by one plus π to the power of π₯ with respect to π₯.
In this question, weβre asked to evaluate the indefinite integral of the quotient of two functions involving the exponential function, π to the power of π₯. And this is a very difficult integral to evaluate directly. So weβre going to need to use one of our rules for integration to help us evaluate this integral.
To help us evaluate this integral, we can recall the indefinite integral of π prime of π₯ divided by π of π₯ with respect to π₯ is the natural logarithm of the absolute value of π of π₯ plus the constant of integration πΆ. And this is a useful result. If we call the denominator of the quotient in our integrand one plus π to the power of π₯ π of π₯, then we can differentiate this. And we find that π prime of π₯ is equal to π to the power of π₯. And we can see this appears in the numerator of our integrand. So we can use this to evaluate part of this integral.
However, we do need to be careful with how we choose to split up this integrand. For example, if we were to split up our integrand over the two terms in our numerator, then we would get one term, two divided by one plus π to the power of π₯, and another term, three π to the power of π₯ divided by one plus π to the power of π₯. We can integrate the second term by using our integral rule. However, we donβt know how to integrate two divided by one plus π to the power of π₯ with respect to π₯. So instead of splitting our integral over the two terms in our numerator, letβs instead see how many times the denominator goes into the numerator.
The easiest way to do this is to rewrite our numerator in terms of multiples of one plus π to the power of π₯. We can see that two times one plus π to the power of π₯ is two plus two π to the power of π₯. If we add π to the power of π₯ to this, we get the same numerator as our integrand. Therefore, weβve rewritten our integral as the integral of two times one plus π to the power of π₯ plus π to the power of π₯ all divided by one plus π to the power of π₯ with respect to π₯.
And now we can split this integrand over the two terms in the numerator. This gives us the integral of two times one plus π to the π₯ divided by one plus π to the π₯ with respect to π₯ plus the integral of π to the π₯ divided by one plus π to the π₯ with respect to π₯. And now this is written in the form of two integrals which we can evaluate.
Letβs start with evaluating the first integral. We can start by canceling the shared factor of one plus π to the power of π₯ in the numerator and denominator. This gives us the integral of two with respect to π₯. And we know that this is equal to two π₯ plus the constant of integration. However, we donβt need to add our constant of integration until the end since weβre evaluating the sum of two integrals.
We can evaluate our second integral by using our integral rule. If π of π₯ is equal to one plus π to the power of π₯, then π prime of π₯ is π to the power of π₯, which is the numerator of this expression. Therefore, the integral of this expression is the natural logarithm of the absolute value of one plus π to the power of π₯ plus our constant of integration πΆ. And this gives us our final answer.
The integral of two plus three π to the power of π₯ all divided by one plus π to the power of π₯ with respect to π₯ is two π₯ plus the natural logarithm of the absolute value of one plus π to the power of π₯ plus the constant of integration πΆ.
