### Video Transcript

For the given function π of π₯
equals the natural log of one plus two π₯, find a power series representation for π
by integrating the power series for π prime.

We donβt have a nice power series
expansion for the function π of π₯ equals the natural log of one plus two π₯. But we should notice that the
derivative of π, π prime of π₯, is two over one plus two π₯. So weβre going to start with an
equation that weβve seen before. That is, one over one minus π₯ is
equal to one plus π₯ plus π₯ squared plus π₯ cubed, and so on. And we write this as the sum of π₯
to the πth power for values of π between zero and β.

We write our derivative as two
times one over one plus two π₯. And then weβre going to use the
equation weβve seen before, replacing π₯ with negative two π₯. And this is because we want it in
the form one minus something. And one minus negative two π₯ gives
us the one plus two π₯ weβre after. We can therefore use this equation
to write one over one plus two π₯ as one plus negative two π₯ plus negative two π₯
squared plus negative two π₯ cubed, and so on. Which can of course be written as
the sum of negative two π₯ to the πth power for values of π between zero and
β.

π prime of π₯ is two times one
over one plus two π₯. So thatβs two times the sum of
negative two π₯ to the πth power for values of π between zero and β. Two is independent of π. So weβre going to take it inside
the sum. And weβre going to rewrite negative
two π₯ as two times negative one times π₯.

We then distribute this exponent
across each term. And we see that we have the sum of
two times two to the πth power times negative one to the πth power times π₯ to the
πth power. Two times two to the πth power can
be written as two to the power of π plus one. And now we have our expression for
π prime of π₯.

Weβre trying to find a power series
representation for π. So we recall that we can achieve
this by integrating our expression for π prime of π₯. And we can do so by integrating
each individual term in the series. Thatβs called term-by-term
integration. Here thatβs the integral of the sum
of two to the power of π plus one times negative one to the πth power times π₯ to
the πth power between π equals zero and β with respect to π₯.

And then of course since weβre
dealing with a power series, we can write this as the sum of the integrals. So how are we going to integrate
two to the power of π plus one times negative one to the πth power times π₯ to the
πth power? Well, no matter the value of π,
two to the power of π plus one times negative one to the πth power is a
constant. This means we can take it outside
of the integral and focus on integrating π₯ to the πth power itself.

Now when we integrate π₯ to the
πth power, we know that π is positive. So we simply add one to the
exponent and then divide by that new number. So we get π₯ to the power of π
plus one over π plus one. And of course we had that constant
of integration πΆ, which is outside the summation. So how do we find that constant of
integration πΆ?

Well, letβs go back to π of
π₯. Weβre told that itβs equal to the
natural log of one plus two π₯. And we know that if we let π₯ be
equal to zero, we get quite a nice value for π of zero. Itβs the natural log of one plus
two times zero, which is the natural log of one, which is of course zero. By replacing π of π₯ with zero and
π₯ with zero, we see that zero is equal to the sum of two to the power of π plus
one times negative one to the πth power plus zero to the power of π plus one over
zero plus one plus πΆ.

Now zero to the power of π plus
one over zero plus one is always zero. And so we have the sum of zeros,
which is zero. And we find the constant of
integration itself is zero. And so by integrating the power
series for π prime, we found a power series representation for π. Itβs the sum of two to the power of
π plus one times negative one to the πth power times π₯ to the power of π plus
one over π plus one.