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.
