# Question Video: Finding the Antiderivative of a Reciprocal Function using Indefinite Integration Mathematics • Higher Education

Find, if possible, an antiderivative 𝐹 of 𝑓(𝑥) = 1/(2𝑥 − 1) that satisfies the conditions 𝐹(0) = 1 and 𝐹(1) = −1.

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### Video Transcript

Find, if possible, an antiderivative capital 𝐹 of lowercase 𝑓 of 𝑥 is equal to one divided by two 𝑥 minus one that satisfies the conditions capital 𝐹 evaluated at zero is one and capital 𝐹 evaluated at one is negative one.

In this question, we’re asked to find an antiderivative of a given reciprocal function that satisfies two conditions; these are called boundary conditions. We can do this by recalling we can determine antiderivatives of functions by using indefinite integration. In particular, the indefinite integral of one divided by two 𝑥 minus one with respect to 𝑥 will give us the most general antiderivative of this function. We know how to integrate the reciprocal function. However, in this case, we can see our denominator is a linear function. So, we’re going to use a 𝑢-substitution.

We’re going to let 𝑢 be equal to the denominator, two 𝑥 minus one. Now, remember to integrate by substitution, we need to find an equation involving the differentials. And we can do this by differentiating both sides of our substitution with respect to 𝑥. Since two 𝑥 minus one is a linear function, its derivative with respect to 𝑥 will be the coefficient of 𝑥, which is two. Now, although d𝑢 by d𝑥 is not a fraction, we can treat it a little bit like a fraction when we’re using integration by substitution. This will allow us to find an equation involving the differentials. This then gives us that a half d𝑢 is equal to d𝑥.

Now, we can use integration by substitution. This will allow us to rewrite our integral. We rewrite the denominator of our integrand as 𝑢, and we replace d𝑥 with a half d𝑢. We get the indefinite integral of one over 𝑢 times a half with respect to 𝑢. And we can simplify our integral slightly by taking the constant factor of a half outside of our integral. This gives us a half times the indefinite integral of one over 𝑢 with respect to 𝑢. We can now evaluate this integral by recalling the integral of one over 𝑥 with respect to 𝑥 is equal to the natural logarithm of the absolute value of 𝑥 plus the constant of integration 𝐶. In our case, our variable is 𝑢. So, we get one-half times the natural logarithm of the absolute value of 𝑢 plus the constant of integration 𝐶.

Well, it’s worth noting we don’t need to multiply our constant by one-half since one-half times a constant is still a constant, so we’ll just call this 𝐶. But remember, our original integrand is in terms of 𝑥 and our original function lowercase 𝑓 of 𝑥 is also in terms of 𝑥. So, we should rewrite our answer to be in terms of the variable 𝑥. We can do this by using our 𝑢-substitution. We get a half times the natural logarithm of the absolute value of two 𝑥 minus one plus 𝐶. And at this point, we might be tempted to call this our function capital 𝐹 and then start substituting 𝑥 is equal to zero and 𝑥 is equal to one and solving for 𝐶. However, while this method would usually work, it won’t work in this case because we’re actually using shorthand notation.

We’re taking the natural logarithm of the absolute value of two 𝑥 minus one. This means that this is a piecewise function. And this makes a lot of sense because to differentiate the natural logarithm of the absolute value of 𝑥, we do this in two parts. The derivative of the natural logarithm of 𝑥 when 𝑥 is greater than zero is one over 𝑥, and the derivative of the natural logarithm of negative 𝑥 is one over 𝑥 when 𝑥 is less than zero. In the same way, our integral result will actually result in a piecewise function, depending on whether our argument is positive or negative.

So, we’re going to need to write this as a piecewise function. We’ll do this by noticing when 𝑥 is greater than one-half, two 𝑥 minus one will be positive, and when 𝑥 is less than one-half, two 𝑥 minus one will be negative. So, if our value of 𝑥 is greater than one-half, then the absolute value of two 𝑥 minus one is just equal to two 𝑥 minus one. And if 𝑥 is less than one-half, then two 𝑥 minus one is negative. So, the absolute value of two 𝑥 minus one will be negative one times two 𝑥 minus one, which is one minus two 𝑥.

This allows us to write our piecewise function capital 𝐹 of 𝑥. It’s equal to a half times the natural logarithm of two 𝑥 minus one plus the constant of integration we’ve called 𝐶 sub one when 𝑥 is greater than one-half. And it’s equal to one-half times the natural logarithm of one minus two 𝑥 plus the constant of integration 𝐶 sub two if 𝑥 is less than one-half. And the important thing to note here is our constants of integration don’t need to be the same. Because remember, when we differentiate this function, we’ll differentiate each subfunction separately over its subdomain. And the derivatives of these constants will always be equal to zero; it doesn’t matter what their values are.

And finally, it may be worth noting 𝑥 is equal to one-half is not included in either subdomain. So, we don’t need to worry about this if we were to differentiate capital 𝐹 of 𝑥. This is simply because one-half is not in the domain of our original function, lowercase 𝑓 of 𝑥. We can now determine the values of 𝐶 sub one and 𝐶 sub two from the question. We’ll start by substituting 𝑥 is equal to zero into this function. To substitute 𝑥 is equal to zero into our piecewise-defined function, we note that zero is less than one-half, so we need to use our second subfunction. We get that capital 𝐹 evaluated at zero is equal to a half times the natural logarithm of one minus two times zero plus 𝐶 sub two. We’re told in the question capital 𝐹 evaluated at zero is equal to one.

Next, one minus two times zero is equal to one, and the natural logarithm of one is just equal to zero. So, this term just simplifies to give us zero. Therefore, we’ve shown that one is equal to 𝐶 sub two. We can now substitute 𝑥 is equal to one into our function capital 𝐹 to determine the value of 𝐶 sub one. Since one is greater than one-half, this is in the first sub domain of our piecewise-defined function. So, capital 𝐹 evaluated at one is one-half times the natural logarithm of two times one minus one plus 𝐶 sub one. We’re told in the question capital 𝐹 evaluated at one is negative one, and we can note two times one minus one is one. And the natural logarithm of one is just zero. So, this simplifies to give us that 𝐶 sub one is equal to negative one.

Now, all we need to do is substitute our values for 𝐶 sub one and 𝐶 sub two into our function capital 𝐹 of 𝑥. This then gives us the following. And it’s worth noting we can technically write our subfunctions in any order; however, usually we write this so the subdomain on the left is at the top. And since 𝑥 being less than one-half starts at the left side of our number line, we’ll write this at the top of our function. And this then gives us our final answer. Capital 𝐹 of 𝑥 will be equal to a half times the natural logarithm of one minus two 𝑥 plus one if 𝑥 is less than one-half. And capital 𝐹 of 𝑥 will be equal to a half times the natural logarithm of two 𝑥 minus one minus one if 𝑥 is greater than one-half.