### Video Transcript

Use partial fractions to evaluate
the integral of ๐ฅ squared plus four ๐ฅ plus one all over ๐ฅ minus one multiplied by
๐ฅ plus one multiplied by ๐ฅ plus three with respect to ๐ฅ.

Weโre told to use partial fractions
to evaluate this integral. So, weโll 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 itโs equal to
some number over ๐ฅ minus one plus some other number over ๐ฅ plus one plus some
other number over ๐ฅ plus three.

What we do in ๐ฅ 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 ๐ฅ squared plus four ๐ฅ plus one. On the right, itโs a little
trickier. When we multiply this first term by
the denominator, the ๐ฅ minus ones cancel. And that means weโre left with ๐ด
multiplied by ๐ฅ plus one multiplied by ๐ฅ plus three.

With our second fraction, ๐ฅ plus
one cancels and weโre left with ๐ต multiplied by ๐ฅ minus one multiplied by ๐ฅ plus
three. And with our last fraction, the ๐ฅ
plus three cancels. So, weโre left with ๐ถ multiplied
by ๐ฅ minus one multiplied by ๐ฅ 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 itโs true for every
value of ๐ฅ. So, we can substitute any values of
๐ฅ into this identity. In fact, we choose to substitute
the roots of the denominator into the identity and use that information to solve for
๐ด, ๐ต, and ๐ถ. Letโs see what this looks like.

One of the roots is ๐ฅ 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
๐ด multiplied by two multiplied by four, which is eight ๐ด. And in fact, the other two terms
disappear because we have ๐ฅ minus one, which we now know is one minus one, which
gives us zero. So, weโre left with six is equal to
eight ๐ด. And we solve this equation for ๐ด
by dividing through by eight. So, ๐ด is equal to six-eighths,
which, of course, simplifies to three-quarters. Weโll repeat this with the next
root.

Another root of this equation is ๐ฅ
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 ๐ด and
๐ถ is zero. So, we have on the right-hand side
๐ต 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 ๐ต. And dividing through by negative
four, we see that ๐ต is equal to one-half.

Finally, weโre interested in the
third root. And thatโs when ๐ฅ 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 ๐ด and ๐ต are zero.

So, we have ๐ถ multiplied by ๐ฅ
minus one, which becomes negative three minus one, which is negative four,
multiplied by ๐ฅ plus one negative three plus one, which is negative two. And we can see that negative two is
equal to eight ๐ถ. This time to solve for ๐ถ, we
divide through by eight. And weโre left with ๐ถ is equal to
negative one-quarter.

We now have enough information to
write our integral in partial fraction form. We substitute ๐ด, ๐ต, and ๐ถ into
our earlier equation. And we see that our integral is
equal to the integral of three over four lots of ๐ฅ minus one plus one over two lots
of ๐ฅ plus one minus one over four lots of ๐ฅ plus three. All thatโs 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 ๐ฅ minus one with respect to ๐ฅ and so on. And then we recall this fact. The integral of ๐ dash ๐ฅ over ๐
of ๐ฅ with respect to ๐ฅ is equal to ln of ๐ of ๐ฅ plus ๐. And the integral of one over ๐ฅ
minus one is ln of ๐ฅ minus one plus ๐. And all of thatโs multiplied by
three-quarters.

In fact, weโre going to be dealing
with a few constants of integration here, so letโs call that ๐ one. Integrating one over ๐ฅ plus one
with respect to ๐ฅ, and we get ln of ๐ฅ plus one plus our constant of
integration. And we multiply all of that by
one-half. And finally, we integrate one over
๐ฅ plus three with respect to ๐ฅ and we get ln of ๐ฅ plus three plus another
constant of integration. And all of that, this time, is
multiplied by negative one-quarter.

Letโs simplify this somewhat. We can combine each of these
constants of integration. Letโs call that ๐พ. So, weโre left with three-quarters
of ln of ๐ฅ minus one plus one-half of ln of ๐ฅ plus one minus one-quarter of ln of
๐ฅ plus three plus our new constant of integration ๐พ. And we arenโt required to simplify
this using the laws of logs, so weโre done.