### Video Transcript

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

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

So this means weโve rewritten ๐ฅ
squared minus two ๐ฅ minus eight as ๐ฅ 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 ๐ฅ minus one squared minus nine with respect to ๐ฅ. 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 ๐ก divided by ๐ with respect
to ๐ก is equal to one divided by the square root of ๐ก squared minus ๐ squared.

To write our integrand in this
form, weโll want to use the substitution ๐ข is equal to ๐ฅ minus one. Weโll differentiate both sides of
this substitution with respect to ๐ฅ. We get d๐ข by the d๐ฅ is equal to
one. And we know that d๐ข by d๐ฅ 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 d๐ข is equal to d๐ฅ. This means to evaluate our integral
using our ๐ข substitution, weโll write d๐ฅ with d๐ข and ๐ฅ minus one with ๐ข. This gives us the integral of one
divided by the square root of ๐ข squared minus nine with respect to ๐ข.

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

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