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.
