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

Evaluate โซ (d๐ฅ/โ(๐ฅยฒ โ 2๐ฅ โ 8)).

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Video Transcript

Evaluate the integral of one divided by the square root of ๐ฅ squared minus two ๐ฅ minus eight with respect to ๐ฅ.

The question is asking us to evaluate this integral. This is a very complicated-looking integral. We canโt evaluate this integral directly. In fact, itโ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 weโll try completing the square. To complete the square, weโll start by halving our coefficient of ๐ฅ. We get ๐ฅ minus one all squared. We then see that this is equal to ๐ฅ squared minus two ๐ฅ plus one. But we want negative eight. So we need to subtract nine from both sides of this equation.

So this means weโve rewritten ๐ฅ squared minus two ๐ฅ minus eight as ๐ฅ 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 ๐ฅ minus one squared minus nine with respect to ๐ฅ. We can then notice this is starting to look like one of our derivative rules. We know, for any constant ๐ not equal to zero, the derivative of the inverse cosh of ๐ก divided by ๐ with respect to ๐ก is equal to one divided by the square root of ๐ก squared minus ๐ squared.

To write our integrand in this form, weโll want to use the substitution ๐ข is equal to ๐ฅ minus one. Weโll differentiate both sides of this substitution with respect to ๐ฅ. We get d๐ข by the d๐ฅ is equal to one. And we know that d๐ข by d๐ฅ 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 d๐ข is equal to d๐ฅ. This means to evaluate our integral using our ๐ข substitution, weโll write d๐ฅ with d๐ข and ๐ฅ minus one with ๐ข. This gives us the integral of one divided by the square root of ๐ข squared minus nine with respect to ๐ข.

And we can now see our integrand is exactly in the form of our derivative rule, where instead of ๐ก we have ๐ข and instead of ๐ we have three because three squared is equal to nine. So by using our integral rule, where ๐ก equal to ๐ข and ๐ equal to three, we can rewrite our integrand as the derivative of the inverse cosh of ๐ข over three with respect to ๐ข. So using this, we get the integral with respect to ๐ข of the derivative of the inverse cosh of ๐ข over three with respect to ๐ข. And now, we can see weโre taking the integral with respect to ๐ข of the derivative with respect to ๐ข. And integration and differentiation are opposite processes.

So this just gives us the inverse cosh of ๐ข divided by three plus our constant of integration ๐ถ. So when we integrate this derivative, weโll just get the function plus a constant of integration ๐ถ. This gives us the inverse cosh of ๐ข over three plus ๐ถ. And remember, weโre evaluating the integral in terms of ๐ฅ. So we should give our answer in terms of ๐ฅ. Weโll do this by using our substitution ๐ข is equal to ๐ฅ minus one. And this gives us our final answer. Weโve shown the integral of one divided by the square root of ๐ฅ squared minus two ๐ฅ minus eight with respect to ๐ฅ is equal to the inverse cosh of ๐ฅ minus one divided by three plus ๐ถ.

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