### Video Transcript

Use partial fractions to evaluate
the integral of π₯ squared plus four π₯ plus one all over π₯ minus one multiplied by
π₯ plus one multiplied by π₯ plus three with respect to π₯.

Weβre told to use partial fractions
to evaluate this integral. So, weβll begin by splitting the
fraction into its parts. And we can only do this when the
denominator is in factorised form. The denominator of our fraction is
already in factorised form. So, we can say that itβs equal to
some number over π₯ minus one plus some other number over π₯ plus one plus some
other number over π₯ plus three.

What we do in π₯ is multiply it by
the denominator of our fraction. This will have the effect of
cancelling out each of the denominators. On the left-hand side, we are left
with π₯ squared plus four π₯ plus one. On the right, itβs a little
trickier. When we multiply this first term by
the denominator, the π₯ minus ones cancel. And that means weβre left with π΄
multiplied by π₯ plus one multiplied by π₯ plus three.

With our second fraction, π₯ plus
one cancels and weβre left with π΅ multiplied by π₯ minus one multiplied by π₯ plus
three. And with our last fraction, the π₯
plus three cancels. So, weβre left with πΆ multiplied
by π₯ minus one multiplied by π₯ plus one.

Now at this stage, we can do one of
two things. We could expand all of the brackets
and compare coefficients. But that can be quite a long
process. Instead, we recognise that this is
an identity. And that means itβs true for every
value of π₯. So, we can substitute any values of
π₯ into this identity. In fact, we choose to substitute
the roots of the denominator into the identity and use that information to solve for
π΄, π΅, and πΆ. Letβs see what this looks like.

One of the roots is π₯ is equal to
one. Substituting one into the left-hand
side of our identity gives us one squared plus four multiplied by one plus one,
which is six. And on the right-hand side, we get
π΄ multiplied by two multiplied by four, which is eight π΄. And in fact, the other two terms
disappear because we have π₯ minus one, which we now know is one minus one, which
gives us zero. So, weβre left with six is equal to
eight π΄. And we solve this equation for π΄
by dividing through by eight. So, π΄ is equal to six-eighths,
which, of course, simplifies to three-quarters. Weβll repeat this with the next
root.

Another root of this equation is π₯
is equal to negative one. When we substitute negative one
into the left-hand side of our identity, we get negative one squared plus four
multiplied by negative one plus one, which is negative two. This time the coefficient of π΄ and
πΆ is zero. So, we have on the right-hand side
π΅ is equal to negative one minus one, which is negative two, multiplied by negative
one plus three, which is two. So, negative two is equal to
negative four π΅. And dividing through by negative
four, we see that π΅ is equal to one-half.

Finally, weβre interested in the
third root. And thatβs when π₯ is equal to
negative three. This time on the left-hand side, we
get negative three squared plus four multiplied by negative three plus one, which is
negative two. And this time, the first and second
terms, the coefficient of π΄ and π΅ are zero.

So, we have πΆ multiplied by π₯
minus one, which becomes negative three minus one, which is negative four,
multiplied by π₯ plus one negative three plus one, which is negative two. And we can see that negative two is
equal to eight πΆ. This time to solve for πΆ, we
divide through by eight. And weβre left with πΆ is equal to
negative one-quarter.

We now have enough information to
write our integral in partial fraction form. We substitute π΄, π΅, and πΆ into
our earlier equation. And we see that our integral is
equal to the integral of three over four lots of π₯ minus one plus one over two lots
of π₯ plus one minus one over four lots of π₯ plus three. All thatβs left is to integrate
this.

It can help to split this integral
up and take the constant factor outside each integral sign. So, we get three-quarters of the
integral of one over π₯ minus one with respect to π₯ and so on. And then we recall this fact. The integral of π dash π₯ over π
of π₯ with respect to π₯ is equal to ln of π of π₯ plus π. And the integral of one over π₯
minus one is ln of π₯ minus one plus π. And all of thatβs multiplied by
three-quarters.

In fact, weβre going to be dealing
with a few constants of integration here, so letβs call that π one. Integrating one over π₯ plus one
with respect to π₯, and we get ln of π₯ plus one plus our constant of
integration. And we multiply all of that by
one-half. And finally, we integrate one over
π₯ plus three with respect to π₯ and we get ln of π₯ plus three plus another
constant of integration. And all of that, this time, is
multiplied by negative one-quarter.

Letβs simplify this somewhat. We can combine each of these
constants of integration. Letβs call that πΎ. So, weβre left with three-quarters
of ln of π₯ minus one plus one-half of ln of π₯ plus one minus one-quarter of ln of
π₯ plus three plus our new constant of integration πΎ. And we arenβt required to simplify
this using the laws of logs, so weβre done.