Video: Finding the General Antiderivative of a Rational Function Involving Polynomial Division to Simplify the Fraction

Find the most general antiderivative capital 𝐹(π‘₯) of the function 𝑓(π‘₯) = (25π‘₯Β² + 29)/(5π‘₯Β² + 5).

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Video Transcript

Find the most general antiderivative capital 𝐹 of π‘₯ of the function lowercase 𝑓 of π‘₯ is equal to 25π‘₯ squared plus 29 all divided by five π‘₯ squared plus five.

In this question, we’re given a function lowercase 𝑓 of π‘₯, which we can see is a rational function. And we need to determine the most general antiderivative capital 𝐹 of π‘₯ of this function. To do this, we need to start by recalling that one way of finding the most general antiderivative of a function lowercase 𝑓 of π‘₯ is to find the indefinite integral of that function with respect to π‘₯. So we want to integrate our function lowercase 𝑓 of π‘₯. However, we can see a problem. This is not an easy function to integrate. In fact, we don’t know how to integrate this directly, so we’re going to need to manipulate this into a form which we can integrate.

One thing we can notice about this function is it’s a rational function where both the numerator and the denominator have the same degree. And we know whenever the degree of the denominator of our rational function is smaller than the degree of our numerator, we can divide through by using polynomial long division. However, in this case, we don’t actually need to perform the entire long division because we can see that our leading terms have the same power. So we know when we divide through by these terms, 25π‘₯ squared over five π‘₯ squared is equal to five. So we know we’re going to get five plus some remainder term divided by five π‘₯ squared plus five. And in fact, we can also find this remainder term without performing the long division.

We’ll write five as a fraction using five π‘₯ squared plus five as the denominator. To do this, we need to multiply five π‘₯ squared plus five all through by five. We see that our constant five is equal to 25π‘₯ squared plus 25 all divided by five π‘₯ squared plus five. And we can see this is almost exactly equal to the rational function we started with. The only difference is in our numerator we need 29 instead of 25. So we need to add four into our numerator. Therefore, we were able to rewrite our original function lowercase 𝑓 of π‘₯. And now we’ll see that this is a much easier expression to integrate.

First, we’ve already shown the integral of lowercase 𝑓 of π‘₯ with respect to π‘₯ will be equal to the integral of five plus four divided by five π‘₯ squared plus five with respect to π‘₯. And we know we can evaluate this integral term by term. We know how to integrate five. However, we don’t yet exactly know how to integrate four divided by five π‘₯ squared plus five. To do this, we’re going to need to notice that this is in a useful form. By recalling the following inverse trigonometric integral rule, the integral of π‘Ž divided by π‘₯ squared plus π‘Ž squared with respect to π‘₯ is equal to the inverse tan of π‘₯ over π‘Ž plus the constant of integration 𝐢 as long as our constant π‘Ž is not equal to zero, we can see that the second term in our integrand is very similar to the integrand in this integral rule.

However, there’s a few key differences. First, our coefficient of π‘₯ squared is not equal to one, so we’ll start by fixing this by taking out a factor of five from our denominator. Doing this, we get the integral of five plus four divided by five times π‘₯ squared plus one with respect to π‘₯. And now we notice in the denominator of our second term of our integrand, we’re adding one. And in our integral rule, we want this to be equal to π‘Ž squared. This means in our numerator, we’re going to want the value of π‘Ž, which is going to be equal to one or negative one. So we’re going to want to take this factor of four and the factor of five outside of our integral.

To do this, let’s start by splitting our integral in two so that we get the integral of five with respect to π‘₯ plus the integral of four divided by five times π‘₯ squared plus one with respect to π‘₯. And in our second integral, we’re just going to take out the factor of four over five. And now we can see that our second integral is just the integral of one divided by π‘₯ squared plus one with respect to π‘₯, which is our integral rule with our value of π‘Ž equal to one.

So now we can just evaluate each of these integrals separately. The integral of five with respect to π‘₯ is equal to five π‘₯. Next, by using our integral rule with π‘Ž equal to one, we get four over five times the inverse tan of π‘₯. And remember, we need to add a constant of integration 𝐢. And there’s one final thing we’ll do. Instead of multiplying by four over five, we’re going to write the entire term over five. And of course, this is equal to our antiderivative capital 𝐹 of π‘₯. Therefore, we were able to show the most general antiderivative capital 𝐹 of π‘₯ of the function lowercase 𝑓 of π‘₯ is equal to 25π‘₯ squared plus 29 all divided by five π‘₯ squared plus five is equal to five π‘₯ plus four times the inverse tan of π‘₯ all over five plus the constant of integration 𝐢.

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