### Video Transcript

Find, if possible, an
antiderivative capital πΉ of π of π₯ equals one over two π₯ minus one that
satisfies the conditions capital πΉ of zero is one and capital πΉ of one is negative
one.

Another way of thinking about the
antiderivative is evaluating the indefinite integral. So, what weβre actually going to do
is integrate one over two π₯ minus one with respect to π₯ and consider the
conditions on it in a moment. To integrate one over two π₯ minus
one, weβre going to use a substitution. Weβre going to let π’ be equal to
two π₯ minus one, which means that dπ’ by dπ₯ equals two. Now, we do know that dπ’ by dπ₯ is
not a fraction, but we do treat it a little like one for the purposes of integration
by substitution and say that a half dπ’ equals dπ₯.

Letβs perform some
substitutions. We replace two π₯ minus one with π’
and dπ₯ with a half dπ’. And we can, of course, take out the
constant factor of one-half, and we have half times the integral of one over π’. Well, remember, the integral of one
over π₯ with respect to π₯ is the natural log of the absolute value of π₯. So, we can say that the integral of
one over two π₯ minus one dπ₯ is a half times the natural log of π’ plus π. We replace π’ with two π₯ minus
one, and weβve obtained our antiderivative. Distributing the parentheses, we
see that the antiderivative π is equal to a half times the natural log of two π₯
minus one plus capital πΆ.

Weβre now going to consider
conditions. But we also recall the fact that
two π₯ minus one cannot be equal to zero. So, weβre actually going to have a
piecewise function for our antiderivative. The first part of the function
weβre interested in is when two π₯ minus one is greater than zero. Solving for π₯, and we obtain π₯ to
be greater than one-half. The second is when one minus two π₯
is greater than zero. Solving for π₯ this time, and we
obtain π₯ to be less than one-half. Letβs clear some space and
formalize this.

We currently have that our
antiderivative, capital πΉ of π₯, is equal to a half times the natural log of one
minus two π₯ plus some constant when π₯ is less than one-half, and a half times the
natural log of two π₯ minus one plus some constant when π₯ is greater than one
half. Weβre now going to use our
conditions on the antiderivative to evaluate π. The first is capital πΉ of zero
equals one. So, when π₯ is equal to zero, πΉ is
equal to one.

Since π₯ is less than one-half,
weβre going to use the first part of our function. We substitute π₯ equals zero and
capital πΉ equals one in. And we get one equals a half times
the natural log of one minus two times zero. Well, the natural log of one minus
two times zero is the natural log of one, which is of course zero. So, we obtain π here to be
one.

The second condition we have is
that when π₯ is equal to one, capital πΉ equals negative one. This time π₯ is greater than
one-half, so weβre going to use the second part of the function. We see that negative one is equal
to a half times the natural log of two times one minus one plus π. Once again, the natural log of two
times one minus one is zero. So, we see that π in the second
part is equal to negative one. And so, the antiderivative capital
πΉ does indeed exist. Such that capital πΉ of π₯ is equal
to a half times the natural log of one minus two π₯ plus one for π₯ is less than
one-half and a half times the natural log of two π₯ minus one minus one for π₯ is
greater than one-half.