### Video Transcript

Determine the integral of 12π₯
minus three multiplied by the eighth root of four π₯ minus one with respect to
π₯.

This looks like it could be a
fairly complicated integral. But when weβre trying to integrate
a function within a root, one really helpful technique is integration by
substitution. And when weβre doing integration by
substitution, sometimes there may be more than one substitution that we could
make. Some may be quicker than
others. Weβll take a look at a couple of
different substitutions that we could make for this question.

Firstly, remember that when we make
a substitution for integration, we often call the substitution π’. And we replace some function of π₯
with π’. But we also have to remember that
we need to replace the dπ₯ because our integral should only involve functions of π’
and dπ’. Letβs make the substitution π’
equals four π₯ minus one, in which case dπ’ by dπ₯, the derivative of π’ with
respect to π₯, is equal to four because four π₯ differentiates to four and negative
one differentiates to zero.

Now, remember that dπ’ by dπ₯ is
not a fraction. But when weβre manipulating in this
way, we do treat it a little bit like a fraction. So by rearrangement, dπ’ is equal
to four dπ₯. And dividing through by four gives
us that dπ₯ is equal to one over four dπ’. So we now know that we can replace
four π₯ minus one with π’ and we can replace dπ₯ with one over four dπ’. But we also need to replace 12π₯
minus three with some function of π’.

Notice that if we take π’ equals
four π₯ minus one and multiply it by three, we have that three π’ equals 12π₯ minus
three. So we know that we can replace 12π₯
minus three in our integral with three π’. So by our substitution, the
integral becomes the integral of three π’ multiplied by the eighth root of π’
multiplied by one over four dπ’.

This is a little messy, so letβs
tidy this integral up a little bit. We can multiply the three with the
one over four to give us three over four. And then remember that we can
actually bring constants out to the front of the integral. So we end up with three over four
multiplied by the integral of π’ multiplied by the eighth root of π’ with respect to
π’.

To simplify this further, we
remember that the πth root of π’ is the same as π’ to the one over π power. So we can rewrite this as three
over four multiplied by the integral of π’ multiplied by π’ to the one over eight
power with respect to π’. Then we can use the fact that π’ to
the π-power multiplied by π’ to the π-power is the same as π’ to the π add π
power. So if we think of π’ on its own as
π’ to the power of one, then π’ to the one power add π’ to the one over eight power
is the same as π’ to the nine over eight power.

So now weβve tidied this up, letβs
integrate. Remember, to integrate this, we
follow the rule of adding one to the power and then dividing by the new power. Adding one to the power gives us π’
to the 17 over eight power. And then we divide by this new
power, which is the same as multiplying by one over 17 over eight. But one over 17 over eight is just
the same as eight over 17. And because this is indefinite
integration, which means we havenβt been given any limits of integration, weβve got
to remember to add a constant of integration πΆ.

Now, to simplify, letβs multiply
three over four by eight over 17. This gives us 24 over 68. But actually, we can divide both
the numerator and the denominator by four to simplify this fraction. It simplifies to six over 17 π’ to
the 17 over eight power plus πΆ.

Now, weβre not completely done with
the question yet because we made the substitution π’ equals four π₯ minus one. So weβve now got to replace π’ with
four π₯ minus one. And that gives us our final
answer.

But now letβs see, just for
comparison, a different substitution we couldβve used to answer this question. Letβs say we use the substitution
π’ equals the eighth root of four π₯ minus one. Then, by raising both sides to the
power of eight, we have that π’ to the eighth power equals four π₯ minus one. We can rearrange this to get that
four π₯ equals π’ to the eighth power add one and then divide through by four to get
that π₯ equals one over four π’ to the eighth power add one over four. The reason that weβve solved for π₯
here is that it makes it a little bit easier to replace dπ₯. Thatβs because, from here, we can
differentiate π₯ with respect to π’ to get that dπ₯ by dπ’ is equal to eight over
four π’ to the seventh power.

This is because we take one over
four π’ to the eighth power, differentiate it with respect to π’ by multiplying by
the power, which is eight, and then subtracting one from the power. And that gives us eight over four
π’ to the seventh power. Of course, one over four is a
constant. So this differentiates to zero. Note that eight over four
simplifies to two. And as weβve already said, dπ₯ by
dπ’ is not a fraction. But we do treat it a little bit
like a fraction when weβre manipulating it in this way. So we can rearrange this to give us
that dπ₯ equals two π’ to the seventh power dπ’.

So now we know that we can replace
the eighth root of four π₯ minus one with the substitution π’. And we can replace dπ₯ with two π’
to the seventh power dπ’. But weβve also got to replace the
12π₯ minus three. Remember that we said π’ to the
eighth power is the same as four π₯ minus one. So three π’ to the eighth power is
equal to three lots of four π₯ minus one, which is 12π₯ minus three. And thatβs what weβre trying to
replace.

So now we know that we can replace
everything in this integral in terms of π’. So letβs go ahead and make these
substitutions. We have the integral of three π’ to
the eighth power multiplied by π’. And then we replace dπ₯ with two π’
to the seventh power dπ’. So our integral is the integral of
three π’ to the eighth power multiplied by π’ multiplied by two π’ to the seventh
power with respect to π’. We can simplify this integral by
multiplying together three with two. And then we can simplify π’ to the
eighth power multiplied by π’ multiplied by π’ to the seventh power by remembering
the rule for multiplying indices together. We can simply add the powers
together. And this gives us π’ to the 16th
power. So our integral is six π’ to the
16th power with respect to π’.

Now we can integrate. Remember, we do this by adding one
to the power and then dividing by the new power. Adding one to the power gives us π’
to the 17th power. And then dividing by that new power
means that we now have six over 17 multiplied by π’ to the 17th power. And because this is indefinite
integration, meaning we donβt have any limits of integration, weβve got to remember
to add a constant of integration πΆ.

Now, remember that we said π’ is
equal to the eighth root of four π₯ minus one. So we can replace π’ with the
eighth root of four π₯ minus one. Now, remember before, we said that
the πth root of something was the same as raising it to the power of one over
π. So the eighth root of four π₯ minus
one is the same as four π₯ minus one to the power of one over eight. And from here, we can tidy this up
a little bit.

If we recall the fact that π
raised to the π-power and then raised to the π-power is the same as π raised to
the π multiplied by π power, this tells us that we can actually just multiply
these two powers together. And so this gives us six over 17
multiplied by four π₯ minus one raised to the power of 17 over eight plus πΆ. So we see that we get exactly the
same answer. Itβs just using a slightly
different substitution.