Question Video: Finding the Integration of a Rational Function by Using Integration by Substitution Mathematics

Video Transcript

Determine the indefinite integral of 𝑥 minus 11 over 𝑥 minus two to the ninth power with respect to 𝑥.

This is not an expression that’s nice to integrate using our standard rules for finding the antiderivative. And so, instead, we’re going to use integration by substitution. The substitution rule says that if 𝑢 equals 𝑔 of 𝑥 is a differentiable function whose range is an interval 𝐼 and 𝑓 is continuous on that interval. Then the integral of 𝑓 of 𝑔 of 𝑥 times 𝑔 prime of 𝑥 with respect to 𝑥 is equal to the integral of 𝑓 of 𝑢 with respect to 𝑢. Now, we can see from our formula that we try to choose 𝑢 to be some factor of the integrand whose derivative also occurs or be at some scalar multiple of it.

Now, if that’s not possible, we try choosing 𝑢 to be a more complicated part of the integrand. This might be the inner function in a composite function. Here, the inner part of the composite function we’re going to choose is 𝑥 minus two, so we’ll let 𝑢 be equal to 𝑥 minus two. We know that the derivative of 𝑢 with respect to 𝑥 is simply one. And whilst d𝑢 by d𝑥 is absolutely not a fraction, we treat it a little like one in this process. And we can say that this is equivalent to saying that d𝑢 is equal to d𝑥. So far, we can replace 𝑥 minus two with 𝑢 and d𝑥 with d𝑢.

We do have a bit of a problem, though. What are we going to do with the numerator of our fraction, with 𝑥 minus 11? Well, if we look back to our substitution, 𝑢 equals 𝑥 minus two, we see if we subtract nine from both sides of this equation, we end up with 𝑥 minus 11 on the right-hand side. So, this means that 𝑢 minus nine equals 𝑥 minus 11. And so, we’re going to need to integrate 𝑢 minus nine over 𝑢 to the ninth power with respect to 𝑢. We’ll begin by writing the integrand as 𝑢 over 𝑢 to the ninth power minus nine over 𝑢 to the ninth power. And then we rewrite each expression a little further. We know that 𝑢 to the power of one divided by 𝑢 to the power of nine is 𝑢 to the power of negative eight. And negative nine over 𝑢 to the ninth power is negative nine 𝑢 to the power of negative nine.

Next, we recall that we can integrate a power term whose exponent is not equal to negative one by adding one to that exponent and then dividing by that new value. It’s worth being a little bit careful here because when we add one to negative eight, we get negative seven. A common mistake is to think that we end up with negative nine. So, we obtain 𝑢 to the power of negative seven over negative seven minus nine 𝑢 to the power of negative eight over negative eight plus 𝐶. We rewrite 𝑢 to the negative seven as one over 𝑢 to the power of seven and 𝑢 to the power of negative eight as one over 𝑢 to the power of eight.

But of course, we were looking to integrate our expression with respect to 𝑥. So, we’re going to go back to our substitution and we’re gonna replace each instance of 𝑢 with 𝑥 minus two. And so, we’ve determined the integral of 𝑥 minus 11 over 𝑥 minus two to the ninth power with respect to 𝑥. It’s negative one over seven times 𝑥 minus two to the seventh power plus nine over eight times 𝑥 minus two to the eighth power plus 𝐶.