### Video Transcript

Determine the indefinite integral
of two π to the power of π₯ times π₯ over three times π₯ plus one all squared,
evaluated with respect to π₯.

It might not be instantly obvious
how we are going to evaluate this. However, thereβs no clear
substitution we can make. And itβs certainly not something we
can evaluate in our heads. So that tells us we might need to
use integration by parts. Remember, this says the integral of
π’ times dπ£ by dπ₯ is equal to π’π£ minus the integral of π£ times dπ’ by dπ₯. Now, if we compare this formula to
our integrand, we see that we need to decide which function is π’. And which function is dπ£ by
dπ₯. So how do we decide this? Well, our aim is to make sure that
the second integral we get over here is a little simpler. We therefore want π’ to be a
function that either becomes simpler when differentiated or helps us to simplify the
integrand when multiplied by this π£. Now it isnβt hugely obvious what we
should choose π’ to be. So a little trial and error might
be in order.

Letβs rewrite our integrand as
two-thirds π₯ times π to the power of π₯ times one over π₯ plus one squared. And in fact, letβs take the
constant factor of two-thirds outside of the integral. Letβs try π’ equals π₯ times π to
the power of π₯ and dπ£ by dπ₯ as being one over π₯ plus one all squared, which Iβve
written as π₯ plus one to the power of negative two. To find dπ’ by dπ₯, weβre going to
need to use the product rule. If we do, we see that dπ’ by dπ₯ is
equal to π₯ times the derivative of π to the power of π₯ plus π to the power of π₯
times the derivative of π₯. Well, the derivative of π to the
power of π₯ is π to the power of π₯. And the derivative of π₯ is
one. So weβve obtained dπ’ by dπ₯ to be
equal to π₯π to the power of π₯ plus π to the power of π₯. We can use the reverse of the chain
rule to work out the antiderivative of π₯ plus one to the power of negative two. Itβs negative π₯ plus one to the
negative one. Rewriting π₯ plus one to the power
of negative one as one over π₯ plus one. And we can substitute everything we
have into our formula for integration by parts.

Now over here, this next step is
important. We factor by π to the power of
π₯. And then we can see we can divide
through by π₯ plus one. And our second integrand becomes
negative π to the π₯. We take out that factor of negative
one. And we know that the integral of π
of the power of π₯ is simply π to the power of π₯. And so we have two-thirds of π₯
times π to the power of π₯ over negative π₯ plus one plus π to the power of π₯
plus π. We rewrite this first fraction
slightly and then multiply the numerator and the denominator of π to the power of
π₯ by π₯ plus one. So that we can add these
fractions. And we see that the sum of these
fractions is negative π₯ times π to the power of π₯ plus π₯ times π to the power
of π₯, which is zero, plus π to the power of π₯ over π₯ plus one.

Our final step is to distribute the
parentheses. And we get two π to the power of
π₯ over three times π₯ plus one. And Iβve changed our constant of
integration to capital πΆ. Since our original constant of
integration was multiplied by two-thirds.