Video Transcript
Evaluate the indefinite integral of
one divided by 𝑥 squared plus four 𝑥 plus five with respect to 𝑥.
We’re given an indefinite integral
which we need to evaluate and we can see that our integrand is a rational
function. And we know several different ways
to evaluate the integral of rational functions. So the first thing we need to do is
decide which method we’re going to use. The first thing we could try is to
check that our integrand is in the form which we already know how to integrate. It’s not in this case; so we’re
going to need to use some kind of manipulation. We could try factoring our
denominator and then using partial fractions. However, if we were to try this, we
would run into a problem. Calculating the discriminant of our
quadratic, we get four squared minus four times five. And if we evaluate this, it’s equal
to negative four. And this quadratic having a
negative discriminant means it has no real roots.
So we can’t factor our
denominator. So we might be stuck out of ideas
of how to manipulate this expression into one which we can integrate. But just because our denominator
has no real roots doesn’t mean we can’t manipulate this. We’ll try using completing the
square. Recall, to complete the square, the
first thing we need to do is divide our coefficient of 𝑥 by two. This gives us 𝑥 plus two all
squared. We can then distribute the square
over our parentheses by either using the FOIL method or binomial expansion. This gives us 𝑥 squared plus four
𝑥 plus four. Of course, in our denominator, we
have 𝑥 plus five instead of 𝑥 plus four. So we need to add an extra constant
one to our expression. So the quadratic in our denominator
is equal to 𝑥 plus two all squared plus one.
So by completing the square, we
were able to rewrite the integral given to us in the question as the integral of one
divided by 𝑥 plus two all squared plus one with respect to 𝑥. And it might be difficult to see
how this helps us evaluate our integral. But now it’s possible to notice our
integral is almost in the form we do know how to integrate. We recall, for any real constant 𝑎
not equal to zero, the integral of 𝑎 divided by 𝑥 squared plus 𝑎 squared with
respect to 𝑥 is equal to the inverse tan of 𝑥 over 𝑎 plus the constant of
integration 𝐶. And this is almost exactly the form
our integral is in. However, instead of 𝑥 squared in
our denominator, we have 𝑥 plus two all squared.
This means we could try integrating
this by substitution. We’ll use the substitution 𝑢 is
equal to 𝑥 plus two. We’ll then differentiate both sides
of this equation with respect to 𝑥; we get d𝑢 by d𝑥 is equal to one. And then we know d𝑢 by d𝑥 is not
a fraction. However, when using integration by
substitution, it can help to treat it a little bit like a fraction. We’ll get the equivalent statement
in terms of differentials d𝑢 is equal to d𝑥. And this is all we need to evaluate
our integral by using our substitution. We replace 𝑥 plus two in our
denominator with 𝑢 and d𝑥 with d𝑢. This gives us the integral of one
over 𝑢 squared plus one with respect to 𝑢.
And now we could see this is in
exactly the same form as our integral rule, with the value of 𝑎 set to be one and
our variable 𝑥 equal to 𝑢. So by setting our value of 𝑎 equal
to one and applying our integral rule, we can evaluate this integral. It’s equal to the inverse tan of 𝑢
plus the constant of integration 𝐶. However, our original integral is
in terms of 𝑥, so we should give our answer in terms of 𝑥 if it’s possible. And we can do this by using our
substitution 𝑢 is equal to 𝑥 plus two. So by using our substitution 𝑢 is
equal to 𝑥 plus two, we get the inverse tan of 𝑥 plus two plus the constant of
integration 𝐶, which is our final answer.
Therefore, we were able to show the
integral of one divided by 𝑥 squared plus four 𝑥 plus five with respect to 𝑥 is
equal to the inverse tan of 𝑥 plus two plus 𝐶.
