### Video Transcript

Indefinite Integrals: Exponential
and Reciprocal Functions

In this video, we will learn how to
find the indefinite integral of exponential and reciprocal functions. Weβll do this by recalling two of
our derivative results involving exponential and reciprocal functions. We can then use these derivative
results to determine antiderivatives of exponential and reciprocal functions. In particular, weβll be able to
find the indefinite integral of exponential functions of the form π to the power of
ππ₯ and reciprocal functions of the form π over π₯. To do this, we recall to determine
the indefinite integral of a function, we want to determine an antiderivative of the
function. And one way of doing this is to
reverse derivative results. For example, we recall if π of π₯
is the exponential function π to the power of π₯, then we know π prime of π₯ is
also π to the power of π₯.

This tells us two things. First, it tells us exactly what it
says. We can differentiate π to the
power of π₯ to just get π to the power of π₯. However, this also tells us another
result: π to the power of π₯ is an antiderivative of π to the power of π₯. Therefore, π to the power of π₯
plus πΆ is the most general antiderivative of this exponential function. In other words, the indefinite
integral of π to the power of π₯ with respect to π₯ is π to the power of π₯ plus
the constant of integration πΆ. At this point, we could move on to
reciprocal functions. However, thereβs two different
forms of exponential functions we want to cover first.

Letβs consider exponential
functions of the form π to the power of π times π₯, where π is a real
constant. And we know how to differentiate
functions like this. We recall π prime of π₯ will be
equal to π times π to the power of ππ₯. We just multiplied by the
coefficient of π₯. And just as we did above, we could
turn this into an integral result. The integral of ππ to the power
of ππ₯ with respect to π₯ is π to the power of ππ₯ plus πΆ. However, itβs more common to just
see functions of the form π to the power of ππ₯. So, we want to remove this value of
π. We do this by multiplying our
original function by one over π, where we assume our value of π is nonzero. Then, we see that π prime of π₯ is
one over π times π times π to the power of ππ₯, which is just π to the power of
ππ₯.

Therefore, weβve shown one over π
times π to the power of ππ₯ is an antiderivative of π to the power of ππ₯,
provided π is nonzero, which also means that weβve shown the indefinite integral of
π to the power of ππ₯ with respect to π₯ is one over π times π to the power of
ππ₯ plus πΆ, provided π is nonzero.

Thereβs one more type of
exponential function we want to look at before we move on to reciprocal
functions. So far, all of our exponential
functions have had a base of π; however, we can consider what happens if we have a
general base. We want to consider the exponential
function π of π₯ is equal to π to the power of π₯, where our base needs to be
positive and not equal to one. We can recall π prime of π₯ will
be equal to π to the power of π₯ multiplied by the natural logarithm of π. Just as we did above, we can write
this into an integral result. Weβve shown that π to the power of
π₯ is an antiderivative of π to the power of π₯ times the natural logarithm of
π.

Therefore, the integral of π to
the power of π₯ times the natural logarithm of π with respect to π₯ is equal to π
to the power of π₯ plus the constant of integration πΆ, provided π is positive and
not equal to one. And since π is not equal to one,
we can divide through by the natural logarithm of π. This gives us the indefinite
integral of π to the power of π₯ with respect to π₯ is equal to π to the power of
π₯ divided by the natural logarithm of π plus the constant of integration πΆ. Well, itβs worth noting we can
write πΆ in both of these cases because πΆ is just a constant. Dividing through by the natural
logarithm of π still leaves this as a constant, so we can just call this πΆ in both
cases. Letβs now move on to the final
example of reciprocal functions.

To determine the indefinite
integral of a reciprocal function, we want a function which differentiates to give
us the reciprocal function. And we can do this by recalling if
π of π₯ is the natural logarithm of the absolute value of π₯, then π prime of π₯
will be equal to one over π₯. Therefore, the natural logarithm of
the absolute value of π₯ is an antiderivative of one over π₯, which tells us the
indefinite integral of one over π₯ with respect to π₯ is the natural logarithm of
the absolute value of π₯ plus πΆ. And we can generalize this result
slightly by multiplying our original function through by π.

If π of π₯ is π times the natural
logarithm of the absolute value of π₯, then π prime of π₯ is π over π₯, which then
tells us for any real constant π the integral of π over π₯ with respect to π₯ is
π times the natural logarithm of the absolute value of π₯ plus πΆ. Letβs now look at some examples
where we use these results to evaluate the integral of more complicated
functions.

Determine the indefinite integral
of eight times π to the power of three π₯ minus π to the power of two π₯ plus nine
divided by seven π to the power of π₯ with respect to π₯.

In this question, weβre asked to
evaluate the indefinite integral of an exponential expression. And at first, it might look very
difficult to evaluate this integral. For example, we might be
considering a π’-substitution. However, we should always check if
we can simplify our integrand. And we can do this by dividing
every term in the numerator by the denominator, where we remember for each term in
the numerator, we just need to subtract the exponents of π. This gives us the indefinite
integral of eight over seven times π to the power of two π₯ minus one-seventh π to
the power of π₯ plus nine over seven times π to the power of negative π₯ with
respect to π₯.

Now, we can see we know how to
integrate each term in the integrand separately. So, weβre going to split this into
three separate integrals. And we can do this because we
recall the integral of the sum of functions is equal to the sum of their individual
integrals. This gives us the following. We can then simplify each term
separately. Weβll take the constant factor
outside of each integral. This then gives us eight-sevenths
times the indefinite integral of π to the power of two π₯ with respect to π₯ minus
one-seventh the indefinite integral of π to the power of π₯ with respect to π₯ plus
nine-sevenths the integral of π to the power of negative π₯ with respect to π₯.

Weβre now ready to evaluate each of
these indefinite integral separately. We can do this by recalling for any
real constant π not equal to zero, the integral of π to the power of ππ₯ with
respect to π₯ is one over π times π to the power of ππ₯ plus the constant of
integration πΆ. Weβll use this to evaluate each
integral separately. Letβs start with the first
integral, the integral of π to the power of two π₯ with respect to π₯. Our value of π is two. So, we substitute π is two into
our integral rule. We get one-half times π to the
power of two π₯. And we need to multiply this by
eight over seven. And itβs worth noting since weβre
evaluating the sum of multiple indefinite integrals, weβll get a constant of
integration for each of these integrals. We can combine all of these
constants into one constant at the end of our expression. So, we donβt need to worry about
adding πΆ until the end.

Letβs now evaluate our second
integral. We know π to the power of π₯ is
the same as π to the power of one times π₯. So, our value of π will be
one. Of course, this is somewhat not
necessary since we know the integral of π to the power of π₯ with respect to π₯ is
just π to the power of π₯ with respect to π₯. However, if we did substitute π is
equal to one, we would get one over one times π to the power of one π₯, which is
just π to the power of π₯. So, evaluating our second integral
and multiplying by negative one-seventh, we get negative one-seventh π to the power
of π₯.

Finally, letβs evaluate the third
integral. Our value of π is negative
one. Substituting π is negative one
into our integral result gives us one over negative one times π to the power of
negative π₯. We need to multiply this by nine
over seven. And remember, we also need to add a
constant of integration at the end of this expression. We can then simplify this
slightly. One over negative one is just equal
to negative one. So, instead of adding this term, we
can subtract this term. And we can also notice in the first
term, we have a shared factor of two in the numerator and denominator. We can cancel this to get a factor
of four over seven.

This then gives us our final
answer. The indefinite integral of π to
the power of three π₯ minus π to the power two π₯ plus nine over seven π to the π₯
with respect to π₯ is equal to four over seven times π to the power of two π₯ minus
π to the power of π₯ over seven minus nine-sevenths π to the power of negative π₯
plus the constant of integration πΆ.

In our previous example, we were
able to rewrite our integrand into a form where we could integrate each term by
using our exponential integral rules. And although it was more difficult,
we also noted we couldβve tried using the π’-substitution. And weβll have a similar fort
process in the next example. We can either attempt to do this by
using a π’-substitution or rewriting our integrand.

Determine the indefinite integral
of two to the power of nine π₯ with respect to π₯.

In this question, weβre asked to
evaluate the integral of an exponential function. And we can note this looks very
similar to one of our integral results. For any positive constant π not
equal to one, the integral of π to the power of π₯ with respect to π₯ is equal to
π to the power of π₯ divided by the natural logarithm of π plus the constant of
integration πΆ. However, in this case, our exponent
is not just equal to π₯; we have nine π₯. So, we canβt just directly apply
this integral result. And there are two ways we can try
and fix this. We could try using the
π’-substitution. π’ is nine π₯. This would then allow us to rewrite
our integral in this form. However, thereβs actually a simpler
method.

We can just apply our laws of
exponents. Two to the ninth power all raised
to the power of π₯ will be equal to two to the power of nine times π₯. So, we can rewrite our integral as
the integral of two to the ninth power all raised to the power of π₯ with respect to
π₯. Then, our value of π is two to the
ninth power. We could now evaluate two to the
ninth power as 512. However, weβre just going to
substitute π is equal to two to the power of nine into our integral result. This then gives us two to the ninth
power all raised to the power of π₯ all divided by the natural logarithm of two to
the ninth power plus the constant of integration πΆ. And this can help us see why itβs
useful to leave this as two to the ninth power rather than evaluate.

We can simplify the numerator by
using our laws of exponents, and we can simplify our denominator by using the power
rule for logarithms. First, in the numerator, two raised
to the power of nine all raised to the power of π₯ will be two to the power of nine
times π₯. Next, in the denominator, the power
rule for logarithms tells us the natural logarithm of two to the ninth power will be
nine times the natural logarithm of two, which then gives us our final answer. The integral of two to the power of
nine π₯ with respect to π₯ is two to the power of nine π₯ divided by nine times the
natural logarithm of two plus the constant of integration πΆ.

Itβs worth noting we can use the
method we used in this question to prove a general integral result. We can apply this method to the
indefinite integral of π to the power of ππ₯ with respect to π₯. We rewrite our integrand as π to
the power of π all raised to the power of π₯ and then use our integral result. Then, we rewrite the numerator by
using our laws of exponents and the denominator by using the power rule for
logarithms. We get π to the power of ππ₯
divided by π times the natural logarithm of π plus the constant of integration
πΆ.

Letβs now see an example of
integrating a reciprocal function.

Determine the indefinite integral
of negative two divided by seven π₯ with respect to π₯.

In this question, weβre asked to
evaluate the indefinite integral of a reciprocal function. And we can do this directly by
recalling one of our integral results. For any real constant π, the
indefinite integral of π over π₯ with respect to π₯ is equal to π times the
natural logarithm of the absolute value of π₯ plus the constant of integration
πΆ. We can just note that our value of
π will be negative two over seven; however, sometimes it can be difficult to see
this. So, instead, weβll just take the
constant factor of negative two over seven outside of our integral. This leaves us with negative two
over seven times the indefinite integral of the reciprocal function, one over π₯
with respect to π₯.

And we know the integral of the
reciprocal function is the natural logarithm of the absolute value of π₯. We can just recall this, or we can
set our value of π equal to one into our integral result. Using either method, weβve shown
the indefinite integral of negative two over seven π₯ with respect to π₯ is negative
two-sevenths times the natural logarithm of the absolute value of π₯ plus πΆ. And itβs worth noting we can always
check our answer by using differentiation. Remember, when weβre finding the
indefinite integral of a function, weβre finding its most general
antiderivative. This means when we differentiate
our function with respect to π₯, we should end up with our integrand.

So, letβs evaluate the derivative
of negative two-sevenths times the natural logarithm of the absolute value of π₯
plus πΆ with respect to π₯. We can start by recalling the
derivative of the natural logarithm of the absolute value of π₯ with respect to π₯
is one over π₯. So, when we differentiate the first
term with respect to π₯, we get negative two over seven multiplied by one over
π₯. The second term is a constant, so
its rate of change with respect to π₯ is zero. This just leaves us with negative
two-sevenths times one over π₯, which we can simplify is negative two over seven
π₯. This is our integrand, so this
confirms that our answer is correct. Therefore, the indefinite integral
of negative two over seven π₯ with respect to π₯ is negative two over seven times
the natural logarithm of the absolute value of π₯ plus πΆ.

Letβs now see an example where we
need to use boundary conditions to determine a specific antiderivative of a given
reciprocal function.

Find, if possible, an
antiderivative capital πΉ of lowercase π of π₯ is equal to one divided by two π₯
minus one that satisfies the conditions capital πΉ evaluated at zero is one and
capital πΉ evaluated at one is negative one.

In this question, weβre asked to
find an antiderivative of a given reciprocal function that satisfies two conditions;
these are called boundary conditions. We can do this by recalling we can
determine antiderivatives of functions by using indefinite integration. In particular, the indefinite
integral of one divided by two π₯ minus one with respect to π₯ will give us the most
general antiderivative of this function. We know how to integrate the
reciprocal function. However, in this case, we can see
our denominator is a linear function. So, weβre going to use a
π’-substitution.

Weβre going to let π’ be equal to
the denominator, two π₯ minus one. Now, remember to integrate by
substitution, we need to find an equation involving the differentials. And we can do this by
differentiating both sides of our substitution with respect to π₯. Since two π₯ minus one is a linear
function, its derivative with respect to π₯ will be the coefficient of π₯, which is
two. Now, although dπ’ by dπ₯ is not a
fraction, we can treat it a little bit like a fraction when weβre using integration
by substitution. This will allow us to find an
equation involving the differentials. This then gives us that a half dπ’
is equal to dπ₯.

Now, we can use integration by
substitution. This will allow us to rewrite our
integral. We rewrite the denominator of our
integrand as π’, and we replace dπ₯ with a half dπ’. We get the indefinite integral of
one over π’ times a half with respect to π’. And we can simplify our integral
slightly by taking the constant factor of a half outside of our integral. This gives us a half times the
indefinite integral of one over π’ with respect to π’. We can now evaluate this integral
by recalling the integral of one over π₯ with respect to π₯ is equal to the natural
logarithm of the absolute value of π₯ plus the constant of integration πΆ. In our case, our variable is
π’. So, we get one-half times the
natural logarithm of the absolute value of π’ plus the constant of integration
πΆ.

Well, itβs worth noting we donβt
need to multiply our constant by one-half since one-half times a constant is still a
constant, so weβll just call this πΆ. But remember, our original
integrand is in terms of π₯ and our original function lowercase π of π₯ is also in
terms of π₯. So, we should rewrite our answer to
be in terms of the variable π₯. We can do this by using our
π’-substitution. We get a half times the natural
logarithm of the absolute value of two π₯ minus one plus πΆ. And at this point, we might be
tempted to call this our function capital πΉ and then start substituting π₯ is equal
to zero and π₯ is equal to one and solving for πΆ. However, while this method would
usually work, it wonβt work in this case because weβre actually using shorthand
notation.

Weβre taking the natural logarithm
of the absolute value of two π₯ minus one. This means that this is a piecewise
function. And this makes a lot of sense
because to differentiate the natural logarithm of the absolute value of π₯, we do
this in two parts. The derivative of the natural
logarithm of π₯ when π₯ is greater than zero is one over π₯, and the derivative of
the natural logarithm of negative π₯ is one over π₯ when π₯ is less than zero. In the same way, our integral
result will actually result in a piecewise function, depending on whether our
argument is positive or negative.

So, weβre going to need to write
this as a piecewise function. Weβll do this by noticing when π₯
is greater than one-half, two π₯ minus one will be positive, and when π₯ is less
than one-half, two π₯ minus one will be negative. So, if our value of π₯ is greater
than one-half, then the absolute value of two π₯ minus one is just equal to two π₯
minus one. And if π₯ is less than one-half,
then two π₯ minus one is negative. So, the absolute value of two π₯
minus one will be negative one times two π₯ minus one, which is one minus two
π₯.

This allows us to write our
piecewise function capital πΉ of π₯. Itβs equal to a half times the
natural logarithm of two π₯ minus one plus the constant of integration weβve called
πΆ sub one when π₯ is greater than one-half. And itβs equal to one-half times
the natural logarithm of one minus two π₯ plus the constant of integration πΆ sub
two if π₯ is less than one-half. And the important thing to note
here is our constants of integration donβt need to be the same. Because remember, when we
differentiate this function, weβll differentiate each subfunction separately over
its subdomain. And the derivatives of these
constants will always be equal to zero; it doesnβt matter what their values are.

And finally, it may be worth noting
π₯ is equal to one-half is not included in either subdomain. So, we donβt need to worry about
this if we were to differentiate capital πΉ of π₯. This is simply because one-half is
not in the domain of our original function, lowercase π of π₯. We can now determine the values of
πΆ sub one and πΆ sub two from the question. Weβll start by substituting π₯ is
equal to zero into this function. To substitute π₯ is equal to zero
into our piecewise-defined function, we note that zero is less than one-half, so we
need to use our second subfunction. We get that capital πΉ evaluated at
zero is equal to a half times the natural logarithm of one minus two times zero plus
πΆ sub two. Weβre told in the question capital
πΉ evaluated at zero is equal to one.

Next, one minus two times zero is
equal to one, and the natural logarithm of one is just equal to zero. So, this term just simplifies to
give us zero. Therefore, weβve shown that one is
equal to πΆ sub two. We can now substitute π₯ is equal
to one into our function capital πΉ to determine the value of πΆ sub one. Since one is greater than one-half,
this is in the first sub domain of our piecewise-defined function. So, capital πΉ evaluated at one is
one-half times the natural logarithm of two times one minus one plus πΆ sub one. Weβre told in the question capital
πΉ evaluated at one is negative one, and we can note two times one minus one is
one. And the natural logarithm of one is
just zero. So, this simplifies to give us that
πΆ sub one is equal to negative one.

Now, all we need to do is
substitute our values for πΆ sub one and πΆ sub two into our function capital πΉ of
π₯. This then gives us the
following. And itβs worth noting we can
technically write our subfunctions in any order; however, usually we write this so
the subdomain on the left is at the top. And since π₯ being less than
one-half starts at the left side of our number line, weβll write this at the top of
our function. And this then gives us our final
answer. Capital πΉ of π₯ will be equal to a
half times the natural logarithm of one minus two π₯ plus one if π₯ is less than
one-half. And capital πΉ of π₯ will be equal
to a half times the natural logarithm of two π₯ minus one minus one if π₯ is greater
than one-half.

Letβs now go over the key points of
this video. First, we were able to reverse our
derivative results involving exponential functions to construct equivalent results
about indefinite integrals. For example, we were able to show
the indefinite integral of π to the power of π₯ with respect to π₯ is π to the
power of π₯ plus πΆ. And for any real constant π not
equal to zero, the indefinite integral of π to the power ππ₯ with respect to π₯ is
equal to one over π times π to the power of ππ₯ plus πΆ.

And finally, for a positive real
constant π not equal to one, the indefinite integral of π to the power of π₯ with
respect to π₯ is π to the power of π₯ divided by the natural logarithm of π plus
πΆ. Secondly, we were able to reverse
another derivative result to show for any real constant π the indefinite integral
of π over π₯ with respect to π₯ is π times the natural logarithm of the absolute
value of π₯ plus πΆ.