# Question Video: Finding the Integration of a Function Involving a Root in Its Denominator by Completing the Square Mathematics • Higher Education

Evaluate ∫ (d𝑥/√(𝑥² − 2𝑥 − 8)).

03:11

### Video Transcript

Evaluate the integral of one divided by the square root of 𝑥 squared minus two 𝑥 minus eight with respect to 𝑥.

The question is asking us to evaluate this integral. This is a very complicated-looking integral. We can’t evaluate this integral directly. In fact, it’s very difficult to see which tool we should use to begin evaluating this integral. Usually, in these cases, the first thing we should try is rewriting our integrand into a form which we can integrate. We see that our integrand contains a quadratic, so we’ll try completing the square. To complete the square, we’ll start by halving our coefficient of 𝑥. We get 𝑥 minus one all squared. We then see that this is equal to 𝑥 squared minus two 𝑥 plus one. But we want negative eight. So we need to subtract nine from both sides of this equation.

So this means we’ve rewritten 𝑥 squared minus two 𝑥 minus eight as 𝑥 minus one all squared minus nine. So we can then use this to rewrite our integrand. We get the integral of one divided by the square root of 𝑥 minus one squared minus nine with respect to 𝑥. We can then notice this is starting to look like one of our derivative rules. We know, for any constant 𝑎 not equal to zero, the derivative of the inverse cosh of 𝑡 divided by 𝑎 with respect to 𝑡 is equal to one divided by the square root of 𝑡 squared minus 𝑎 squared.

To write our integrand in this form, we’ll want to use the substitution 𝑢 is equal to 𝑥 minus one. We’ll differentiate both sides of this substitution with respect to 𝑥. We get d𝑢 by the d𝑥 is equal to one. And we know that d𝑢 by d𝑥 is not a fraction; however, we can treat it a little bit like a fraction. This gives us the equivalent statement in terms of differentials d𝑢 is equal to d𝑥. This means to evaluate our integral using our 𝑢 substitution, we’ll write d𝑥 with d𝑢 and 𝑥 minus one with 𝑢. This gives us the integral of one divided by the square root of 𝑢 squared minus nine with respect to 𝑢.

And we can now see our integrand is exactly in the form of our derivative rule, where instead of 𝑡 we have 𝑢 and instead of 𝑎 we have three because three squared is equal to nine. So by using our integral rule, where 𝑡 equal to 𝑢 and 𝑎 equal to three, we can rewrite our integrand as the derivative of the inverse cosh of 𝑢 over three with respect to 𝑢. So using this, we get the integral with respect to 𝑢 of the derivative of the inverse cosh of 𝑢 over three with respect to 𝑢. And now, we can see we’re taking the integral with respect to 𝑢 of the derivative with respect to 𝑢. And integration and differentiation are opposite processes.

So this just gives us the inverse cosh of 𝑢 divided by three plus our constant of integration 𝐶. So when we integrate this derivative, we’ll just get the function plus a constant of integration 𝐶. This gives us the inverse cosh of 𝑢 over three plus 𝐶. And remember, we’re evaluating the integral in terms of 𝑥. So we should give our answer in terms of 𝑥. We’ll do this by using our substitution 𝑢 is equal to 𝑥 minus one. And this gives us our final answer. We’ve shown the integral of one divided by the square root of 𝑥 squared minus two 𝑥 minus eight with respect to 𝑥 is equal to the inverse cosh of 𝑥 minus one divided by three plus 𝐶.

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