### Video Transcript

Find, if possible, an
antiderivative capital ๐น of lowercase ๐ of ๐ฅ is equal to one divided by two ๐ฅ
minus one that satisfies the conditions capital ๐น evaluated at zero is one and
capital ๐น evaluated at one is negative one.

In this question, weโre 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 ๐ฅ minus one with respect to ๐ฅ 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, weโre going to use a
๐ข-substitution.

Weโre going to let ๐ข be equal to
the denominator, two ๐ฅ 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 ๐ฅ. Since two ๐ฅ minus one is a linear
function, its derivative with respect to ๐ฅ will be the coefficient of ๐ฅ, which is
two. Now, although d๐ข by d๐ฅ is not a
fraction, we can treat it a little bit like a fraction when weโre using integration
by substitution. This will allow us to find an
equation involving the differentials. This then gives us that a half d๐ข
is equal to d๐ฅ.

Now, we can use integration by
substitution. This will allow us to rewrite our
integral. We rewrite the denominator of our
integrand as ๐ข, and we replace d๐ฅ with a half d๐ข. We get the indefinite integral of
one over ๐ข times a half with respect to ๐ข. 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 ๐ข with respect to ๐ข. We can now evaluate this integral
by recalling the integral of one over ๐ฅ with respect to ๐ฅ is equal to the natural
logarithm of the absolute value of ๐ฅ plus the constant of integration ๐ถ. In our case, our variable is
๐ข. So, we get one-half times the
natural logarithm of the absolute value of ๐ข plus the constant of integration
๐ถ.

Well, itโs worth noting we donโt
need to multiply our constant by one-half since one-half times a constant is still a
constant, so weโll just call this ๐ถ. But remember, our original
integrand is in terms of ๐ฅ and our original function lowercase ๐ of ๐ฅ is also in
terms of ๐ฅ. So, we should rewrite our answer to
be in terms of the variable ๐ฅ. We can do this by using our
๐ข-substitution. We get a half times the natural
logarithm of the absolute value of two ๐ฅ minus one plus ๐ถ. And at this point, we might be
tempted to call this our function capital ๐น and then start substituting ๐ฅ is equal
to zero and ๐ฅ is equal to one and solving for ๐ถ. However, while this method would
usually work, it wonโt work in this case because weโre actually using shorthand
notation.

Weโre taking the natural logarithm
of the absolute value of two ๐ฅ 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 ๐ฅ, we do
this in two parts. The derivative of the natural
logarithm of ๐ฅ when ๐ฅ is greater than zero is one over ๐ฅ, and the derivative of
the natural logarithm of negative ๐ฅ is one over ๐ฅ when ๐ฅ 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, weโre going to need to write
this as a piecewise function. Weโll do this by noticing when ๐ฅ
is greater than one-half, two ๐ฅ minus one will be positive, and when ๐ฅ is less
than one-half, two ๐ฅ minus one will be negative. So, if our value of ๐ฅ is greater
than one-half, then the absolute value of two ๐ฅ minus one is just equal to two ๐ฅ
minus one. And if ๐ฅ is less than one-half,
then two ๐ฅ minus one is negative. So, the absolute value of two ๐ฅ
minus one will be negative one times two ๐ฅ minus one, which is one minus two
๐ฅ.

This allows us to write our
piecewise function capital ๐น of ๐ฅ. Itโs equal to a half times the
natural logarithm of two ๐ฅ minus one plus the constant of integration weโve called
๐ถ sub one when ๐ฅ is greater than one-half. And itโs equal to one-half times
the natural logarithm of one minus two ๐ฅ plus the constant of integration ๐ถ sub
two if ๐ฅ is less than one-half. And the important thing to note
here is our constants of integration donโt need to be the same. Because remember, when we
differentiate this function, weโll differentiate each subfunction separately over
its subdomain. And the derivatives of these
constants will always be equal to zero; it doesnโt matter what their values are.

And finally, it may be worth noting
๐ฅ is equal to one-half is not included in either subdomain. So, we donโt need to worry about
this if we were to differentiate capital ๐น of ๐ฅ. This is simply because one-half is
not in the domain of our original function, lowercase ๐ of ๐ฅ. We can now determine the values of
๐ถ sub one and ๐ถ sub two from the question. Weโll start by substituting ๐ฅ is
equal to zero into this function. To substitute ๐ฅ 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 ๐น evaluated at
zero is equal to a half times the natural logarithm of one minus two times zero plus
๐ถ sub two. Weโre told in the question capital
๐น 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, weโve shown that one is
equal to ๐ถ sub two. We can now substitute ๐ฅ is equal
to one into our function capital ๐น to determine the value of ๐ถ sub one. Since one is greater than one-half,
this is in the first sub domain of our piecewise-defined function. So, capital ๐น evaluated at one is
one-half times the natural logarithm of two times one minus one plus ๐ถ sub one. Weโre told in the question capital
๐น 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
๐ถ sub one is equal to negative one.

Now, all we need to do is
substitute our values for ๐ถ sub one and ๐ถ sub two into our function capital ๐น of
๐ฅ. This then gives us the
following. And itโs 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 ๐ฅ being less than
one-half starts at the left side of our number line, weโll write this at the top of
our function. And this then gives us our final
answer. Capital ๐น of ๐ฅ will be equal to a
half times the natural logarithm of one minus two ๐ฅ plus one if ๐ฅ is less than
one-half. And capital ๐น of ๐ฅ will be equal
to a half times the natural logarithm of two ๐ฅ minus one minus one if ๐ฅ is greater
than one-half.