Question Video: Finding the Integration of a Function Involving a Root Function Using Integration by Substitution | Nagwa Question Video: Finding the Integration of a Function Involving a Root Function Using Integration by Substitution | Nagwa

# Question Video: Finding the Integration of a Function Involving a Root Function Using Integration by Substitution Mathematics • Third Year of Secondary School

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Determine β«(12π₯ β 3) multiplied by the eighth root of (4π₯ β 1) dπ₯.

08:09

### 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 weβre trying to integrate a function within a root, one really helpful technique is integration by substitution. And when weβre doing integration by substitution, sometimes there may be more than one substitution that we could make. Some may be quicker than others. Weβll 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’. Letβs 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 weβre 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 letβs 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 πth 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 weβve tidied this up, letβs 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 havenβt been given any limits of integration, weβve got to remember to add a constant of integration πΆ.

Now, to simplify, letβs 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 πΆ.

Now, weβre not completely done with the question yet because we made the substitution π’ equals four π₯ minus one. So weβve now got to replace π’ with four π₯ minus one. And that gives us our final answer.

But now letβs see, just for comparison, a different substitution we couldβve used to answer this question. Letβs 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 weβve solved for π₯ here is that it makes it a little bit easier to replace dπ₯. Thatβs 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 weβve already said, dπ₯ by dπ’ is not a fraction. But we do treat it a little bit like a fraction when weβre 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 weβve 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 thatβs what weβre trying to replace.

So now we know that we can replace everything in this integral in terms of π’. So letβs 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 donβt have any limits of integration, weβve got to remember to add a constant of integration πΆ.

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 πth 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 πΆ. So we see that we get exactly the same answer. Itβs just using a slightly different substitution.

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