### Video Transcript

Use partial fractions to evaluate
the indefinite integral of eight π₯ squared plus eight π₯ plus two over four π₯
squared plus one all squared.

Now, we are actually told how weβre
going to evaluate this integral. Weβre going to write the fraction β
the integrand β in partial faction form. But weβre going to need to be
really careful. And thereβs two reasons for
this. Firstly, one of the factors in our
denominator is a nonfactorable quadratic and, secondly, that quadratic is being
squared. And so, to deal with this, there
are two things we need to do. We separate our fraction into the
sum of two rational functions. Because the denominators of both of
these rational functions are made up of these quadratics, the numerators must be of
the form π΄π₯ plus π΅ and πΆπ₯ plus π·, where π΄, π΅, πΆ, and π· are constants. And because that quadratic
expression is being squared, the numerators are four π₯ squared plus one and four π₯
squared plus one squared.

Now, we need to make the expression
on the right look like that on the left. And to do so, we do something with
this first fraction here. We multiply both the numerator and
denominator of this fraction by four π₯ squared plus one. In doing so, this creates a common
denominator. And then, weβll simply be able to
add the numerators. We get π΄π₯ plus π΅ times four π₯
squared plus one plus πΆπ₯ plus π· all over four π₯ squared plus one squared. Now, this is, of course, equal to
the earlier fraction. And we can see that the
denominators of each of our fractions are equal, which means, in turn, their
numerators must also be equal. That is to say, eight π₯ squared
plus eight π₯ plus two is equal to π΄π₯ plus π΅ times four π₯ squared plus one plus
πΆπ₯ plus π·.

Our job now is to find the values
of the constants π΄, π΅, πΆ, and π·. Now, usually, weβd look to
substitute the zeros of four π₯ squared plus one in. But we donβt have any here. And so, what weβre going to do is
distribute the parentheses on the right-hand side and equate coefficients. Distributing the parentheses and we
get four π΄π₯ cubed plus π΄π₯ plus four π΅π₯ squared plus π΅ plus πΆπ₯ plus π·. Weβre going to begin then by
equating coefficients of π₯ cubed. On the left-hand side, there are no
π₯ cubed terms. So, the coefficient of π₯ cubed is
zero. And on the right, the only
coefficient we have is four π΄. Dividing through by four then and
we find that π΄ itself is equal to zero.

Letβs repeat this process for π₯
squared. This time, on the left, we have
eight. And on the right, itβs four π΅. Dividing both sides of this
equation by four and we find that π΅ must be equal to two. Now, weβll create coefficients of
π₯ or π₯ to the power of one. On the left, thatβs eight, and then
on the right, we now have two terms. We have π΄ and πΆ. But of course, we said that π΄ is
equal to zero. And that means that πΆ must be
equal to eight.

Finally, we equate the coefficients
of π₯ to the power of zero. But of course, those are just the
constant terms. On the left, thatβs two. And on the right, we have π΅ and
π·. Remember, though, we said that π΅
is equal to two. And so, if we subtract two from
both sides of this equation, we find π· is equal to zero. And so, going back to our partial
fraction form, we see that we can write our fraction as zero π₯ plus two over four
π₯ squared plus one plus eight π₯ plus zero over four π₯ squared plus one all
squared.

And weβre now ready to
integrate. Letβs clear some space. We can see that weβre integrating
two over four π₯ squared plus one plus eight π₯ over four π₯ squared plus one all
squared with respect to π₯. Letβs do this term by term. Weβre first going to look at the
expression two over four π₯ squared plus one. Here, itβs really useful to recall
that the derivative of one over one plus π₯ squared is equal to the inverse tan of
π₯. We can take out constant
factors. So, weβll take out that factor of
two. And then, weβre going to write four
π₯ squared as two π₯ all squared. And then, this means when we
integrate one over one plus two π₯ all squared, we get the inverse tan of two
π₯. So, the integral of two times this
is two times the inverse tan of two π₯.

And what about the integral of
eight π₯ over four π₯ squared plus one all squared? Well, if we spot that the numerator
is a scalar multiple of the derivative of the inner function in our denominator,
that means we can use integration by substitution. Now, we let π’ be equal to that in
a function, four π₯ squared plus one. Then, dπ’ by dπ₯, the derivative of
this expression with respect to π₯, is eight π₯. Now, dπ’ by dπ₯ isnβt a fraction,
but we do treat it a little like one. And we can write this as dπ’ equals
eight π₯ dπ₯. Then, we replace eight π₯ dπ₯ with
dπ’ and of course four π₯ squared plus one with π’. So, we need to integrate one over
π’ squared with respect to π’ or π’ to the power of negative two. To integrate this, we add one to
the power and divide by that new value. So, itβs π’ to the power of
negative one over negative one, which is negative one over π’.

Our final step with this
substitution is to replace π’ with four π₯ squared plus one. And so, we see that the integral of
eight π₯ over four π₯ squared plus one all squared is negative one over four π₯
squared plus one. This is an indefinite integral. So, we add our constant of
integration. And we see that the integral of
eight π₯ squared plus eight π₯ plus two all over four π₯ squared plus one all
squared is two times the inverse tan of two π₯ minus one over four π₯ squared plus
one plus our constant of integration π.