### Video Transcript

Integrals involving logarithmic
functions. In this video, we will learn how to
use integration by substitution on function of the form π prime of π₯ over π of
π₯. Weβll be looking at some examples
to see the kind of integrals we can use this method on. Now, letβs start by considering the
following integral. And thatβs the indefinite integral
of π prime of π₯ over π of π₯ with respect to π₯. And weβre going to try to solve
this integral. And we can in fact do this with a
substitution. Letβs let π’ equal π of π₯. And we can differentiate π’ with
respect to π₯. And this gives us dπ’ by dπ₯ is
equal to π prime of π₯. Where the prime denotes a
differentiation with respect to π₯. And this tells us that dπ’ is equal
to π prime of π₯ dπ₯.

Now we can rewrite our integral as
the integral of one over π of π₯ multiplied by π prime of π₯ dπ₯. Weβre now ready to perform our
substitution. We will substitute in π’ for π of
π₯ and dπ’ for π prime of π₯ dπ₯. So you obtain that our integral is
equal to the integral of one over π’ with respect to π’. And this is an integral which we
know how to solve. We know that the integral of one
over π₯ with respect to π₯ is equal to the natural logarithm of the absolute value
of π₯ plus π. Therefore, our integral is equal to
the natural logarithm of the absolute value of π’ plus our integration constant
π. And here, we can substitute our
value of π’ back into our equation. And this leads us to our result,
which is that the indefinite integral of π prime of π₯ over π of π₯ with respect
to π₯ is equal to the natural logarithm of the absolute value of π of π₯ plus
π.

Letβs now look at an example of how
this works.

Determine the indefinite integral
of two π₯ plus one over π₯ squared plus π₯ minus seven with respect to π₯.

When we look at the integrand of
this integral, we can notice that the numerator looks a lot like the derivative of
the denominator. Letβs quickly check this. We can call the denominator π of
π₯. So π of π₯ is equal to π₯ squared
plus π₯ minus seven. And now we can differentiate
this. Using the power rule on each term,
we obtain that π prime of π₯ is equal to two π₯ plus one. Which is equal to the numerator of
our fraction. Therefore, our integrand is of the
form π prime of π₯ over π of π₯. And we in fact have a rule for
integrating functions of this form. It tells us that the integral of π
prime of π₯ over π of π₯ with respect to π₯ is equal to the natural logarithm of
the absolute value of π of π₯ plus π. Using this rule, we obtain that the
integral of two π₯ plus one over π₯ squared plus π₯ minus seven with respect to π₯
is equal to the natural logarithm of the absolute value of π₯ squared plus π₯ minus
seven plus our constant of integration π.

In the next example, we will see
how this method can be very useful when integrating certain trigonometric
functions.

Determine the indefinite integral
of cot π₯ with respect to π₯.

Now we know that cot π₯ can be
written as cos of π₯ over sin of π₯. Therefore, we can rewrite our
integral as the integral of cos π₯ over sin π₯ with respect to π₯. Next, weβll be using the fact that
the differential of sin π₯ with respect to π₯ is equal to cos π₯. And so if we let our denominator
sin π₯ be equal to π of π₯, then our numerator, cos of π₯, will be equal to π
prime of π₯. And here, we can see that our
integral is of the form of the integral of π prime of π₯ over π of π₯ with respect
to π₯.

And we know a formula for solving
integrals of this form. And this formula tells us that the
indefinite integral of π prime of π₯ over π of π₯ with respect to π₯ is equal to
the natural logarithm of the absolute value of π of π₯ plus π. If we apply this formula to our
integral with π of π₯ is equal to sin π₯, then we will reach our solution. Which is that the indefinite
integral of cot π₯ with respect to π₯ is equal to the natural logarithm of the
absolute value of sin π₯ plus π.

Now we can in fact take this method
one step further. And consider when our numerator
differs from the differential of the denominator by a constant factor. So if our integrand is equal to π
times π prime of π₯ over π of π₯, where π is a real number. Now, this may appear very
straightforward. Since we can use the fact that we
can factor out a constant from inside an integral. This gives us that our integral is
equal to π multiplied by the natural logarithm of the absolute value of π of π₯
plus π. And now this method seems very
straightforward. Itβs just a slight step up from
what we weβre doing before. However, there are in fact two
potentially quite difficult parts in setting up our integral. So we can apply this formula.

And the first of these two points
is to actually spot that our integral can be in this form. Since itβs not always obvious that
our numerator is some constant multiplied by the differential of the
denominator. Sometimes we may be required to
multiply our integrand by some function over itself, such as β of π₯ over β of
π₯. In order to get it in the form to
use this method. The second part, which is often
slightly less tricky, is to find the value of our constant π, since itβs not always
immediately obvious.

Letβs now look at an example of how
we can use this method.

Determine the indefinite integral
of π₯ squared plus seven over π₯ cubed plus 21π₯ minus five with respect to π₯.

Now the first thing that we notice
is that the numerator of our integrand looks a lot like the differential of the
denominator. Letβs quickly check this. Letβs let π of π₯ be our
denominator. So π of π₯ is equal to π₯ cubed
plus 21π₯ minus five. And now we differentiate. And we get that π prime of π₯ is
equal to three π₯ squared plus 21, which is not equal to our numerator. However, we notice that itβs in
fact out by a factor of three. If we divide π prime of π₯ by
three, then we obtain that one-third π prime of π₯ is equal to π₯ squared plus
seven. And this is now equal to the
numerator of our integrand. We could write our integral in
terms of π of π₯. And we would obtain that it is
equal to the integral of one-third π prime of π₯ over π of π₯ with respect to
π₯.

Now we can use a rule which we know
to integrate integrals of this form. It tells us that the integral of π
multiplied by π prime of π₯ over π of π₯ with respect to π₯ is equal to π
multiplied by the natural logarithm of the absolute value of π of π₯ plus π. So we can apply this formula. Except in our case, π is
one-third. And π of π₯ is equal to π₯ cubed
plus 21π₯ minus five. In doing this, we reach our
solution, which is that the indefinite integral of π₯ squared plus seven over π₯
cubed plus 21π₯ minus five with respect to π₯ is equal to one-third multiplied by
the natural logarithm of the absolute value of π₯ cubed plus 21π₯ minus five plus
π.

In this last example, we saw how we
can find our constant π by inspection. However, this is not always
particularly obvious. Another way in which we can find it
is by using substitution as weβll see in the next example.

Determine the indefinite integral
of 27 sin π₯ plus 21 cos π₯ all over seven sin π₯ minus nine cos π₯ with respect to
π₯.

Here, we notice that the numerator
of the integrand looks as though it could be the differential of the
denominator. Since the differential of sin π₯
with respect to π₯ is equal to cos π₯ and the differential of negative cos π₯ with
respect to π₯ is equal to sin π₯. Now it looks as though the
numerator may differ from the differential of the denominator by a constant
factor. However, we do not know what this
factor is. We can try and find it by using a
substitution. Letβs let π’ be equal to the
denominator of the integrand, so that seven sin π₯ minus nine cos π₯. Now we can differentiate π’ with
respect to π₯. Using the fact that sine
differentiates to cos π₯ and negative cos π₯ differentiates to sin π₯. We obtain that dπ’ by dπ₯ is equal
to seven cos π₯ plus nine sin π₯. This gives us that dπ’ is equal to
nine sin π₯ plus seven cos π₯ dπ₯.

Now letβs rearrange our integral so
we can apply this substitution. We notice that we can factor out a
factor of three from our numerator. And this enables us to write our
integral as the integral of three over seven sin π₯ minus nine cos π₯ multiplied by
nine sin π₯ plus seven cos π₯ dπ₯. And so we can substitute π’ into
the denominator of our fraction. And we can substitute in dπ’ for
nine sin π₯ plus seven cos π₯ dπ₯. Giving us that it is equal to the
integral of three over π’ dπ’, which can be integrated to three multiplied by the
natural logarithm of the absolute value of π’ plus π. For our final step, we simply
substitute back in seven sin π₯ minus nine cos π₯ for π’. This gives us our solution, which
is three multiplied by the natural logarithm of the absolute value of seven sin π₯
minus nine cos π₯ plus π.

Now this method of integration can
be used to integrate many different types of functions. In the next example, weβll see an
integration of a function involving a natural logarithm.

Determine the indefinite integral
of negative three over π₯ multiplied by the natural logarithm of eight π₯ with
respect to π₯.

Now, although this may look like a
tricky integral to evaluate, it is in fact in a form which we know how to
integrate. If we let π of π₯ be equal to the
natural logarithm of eight π₯, then we can differentiate π of π₯ using the fact
that the differential of the natural logarithm of π₯ is one over π₯ in order to find
that π prime of π₯ is equal to one over eight π₯. And then, since this is a composite
function, we have eight π₯ inside the function of the natural logarithm. We mustnβt forget to multiply by
the differential of eight π₯, which is just eight. This is because of the chain
rule. Simplifying, we can obtain that π
prime of π₯ is equal to one over π₯.

Now letβs rewrite our integral. If we multiply the numerator and
denominator of our fraction by one over π₯, then we can rewrite our integral as the
integral of negative three over π₯ over the natural logarithm of eight π₯ with
respect to π₯. And now we can factor the negative
three in the numerator. And once weβve reached this stage,
we notice that this is in a form which we know how to integrate. Since itβs of the form the integral
of π multiplied by π prime of π₯ over π of π₯ dπ₯. Where our π of π₯ is the natural
logarithm of eight π₯. And our π prime of π₯ is one over
π₯. Therefore, our value of π is
negative three. Now we know that this integral
evaluates to π multiplied by the natural logarithm of the absolute value of π of
π₯ plus π. And we can simply substitute in the
values of π and π of π₯ to find our solution. Which is negative three multiplied
by the natural logarithm of the absolute value of the natural logarithm of eight π₯
plus π.

In our final example, weβll be
seeing an integration of a trigonometric function where weβre required to multiply
our integrand by a fraction of a function over itself.

Determine the indefinite integral
of two multiplied by csc of seven π₯ with respect to π₯.

Now, this is a really tricky
function to integrate. It requires us to really know our
derivatives of trigonometric functions. Now, since weβre trying to
determine the indefinite integral of two csc of seven π₯, letβs write down what the
derivative of csc of seven π₯ is. It is equal to negative seven csc
of seven π₯ multiplied by cot of seven π₯. Now what weβre aiming to do at this
stage is to try and get our integrand in the form of π multiplied by π prime of π₯
over π of π₯. Since we know how to integrate this
and in order to try and get our integral in this form. Weβre going to try and multiply the
integrand by a fraction consisting of a function over itself, which is of course
equal to one. The difficult part is trying to
find a function π of π₯. So our integral will end up in the
form which we know how to integrate.

Letβs see what happens if we let π
of π₯ be equal to csc of seven π₯. We multiply our integrand by csc of
seven π₯ over csc of seven π₯. And we obtain the integral of two
csc squared of seven π₯ over csc of seven π₯ with respect to π₯. However, this is clearly not in the
form required. However, it does give us a hint
since we have a csc squared of seven π₯ in the numerator. And we in fact know another
trigonometric function which would differentiate to give us a multiple of csc
squared of seven π₯. And this is cot of seven π₯. The derivative of cot of seven π₯
is negative seven csc squared of seven π₯.

Now itβs at this stage where we
might start to be able to spot what our π of π₯ is. When we multiply by π of π₯ over
π of π₯ in our integrand, we are always gonna have that factor of csc of seven π₯,
which is originally in the integrand. If we factor out a factor of csc of
seven π₯ from our two differentials, then weβre left with a cot of seven π₯
multiplied by a constant and a csc of seven π₯ multiplied by the same constant. Now, this is very important since
the functions which weβre differentiating are csc of seven π₯ and cot of seven
π₯.

And so maybe, for our next π of
π₯, we can try adding these two functions together. But first, letβs make a quick note
what the differential of the sum of these two functions is. Using the fact that the
differential of a sum of functions is equal to the sum of the differentials of the
functions. And by keeping the csc of seven π₯
term factored. We have that the differential of
csc of seven π₯ plus cot of seven π₯ is equal to csc of seven π₯ multiplied by
negative seven cot of seven π₯ minus seven csc of seven π₯.

So now we can try π of π₯ being
equal to csc of seven π₯ plus cot of seven π₯. We have the integral of two csc of
seven π₯ multiplied by csc of seven π₯ plus cot of seven π₯ over csc of seven π₯
plus cot of seven π₯ with respect to π₯. And we can rewrite this with the
two csc of seven π₯ in the numerator. Now we can notice that our integral
is very nearly in the form which we require. If we let π of π₯ be equal to π
of π₯, then we can see that the denominator of our integrand is π of π₯. And if we compare the numerator
with the differential of π of π₯, then we can see that itβs very, very similar. The only two slight differences is
that thereβs this constant factor of two in the numerator of the integrand and a
factor of negative seven in the differential. And so we can come to the
conclusion that the numerator of our integrand is equal to two over negative seven
times by π prime of π₯.

We can now more clearly see that
our integral is in fact of the form which we require. In our case, π is equal to
negative two over seven. And π of π₯ is equal to csc of
seven π₯ plus cot of seven π₯. And so we can apply the
formula. Here, we reach our solution, which
is that the indefinite integral of two csc of seven π₯ with respect to π₯ is equal
to negative two-sevenths times the natural logarithm of the absolute value of csc of
seven π₯ plus cot of seven π₯ plus π.

In this last example, we saw how we
can sometimes find the integral of a difficult function using this method by
multiplying by π of π₯ over π of π₯ for some function π of π₯. And the most difficult part of this
method is working out what π of π₯ needs to be. Knowing the differentials of many
different types of functions is certainly a very useful tool in doing this.

We have now covered a variety of
examples in this video. Letβs recap some of the key
points. Key points. The indefinite integral of π
multiplied by π prime of π₯ over π of π₯ with respect to π₯ is equal to π
multiplied by the natural logarithm of the absolute value of π of π₯ plus π, where
π is a real number. This method can be used to find the
integrals of many different types of functions, including trigonometric ones such as
cot π₯, tan π₯, csc π₯, and so on. Sometimes it is not immediately
obvious that we can use this method to integrate. But multiplying by π of π₯ over π
of π₯ for some π of π₯ may enable us to use it.