### Video Transcript

Determine the integral of negative 12π₯ add 26 multiplied by the square root of negative 12π₯ add 26 with respect to π₯.

One good technique to integrate a function involving a root function is to use integration by substitution. Remember, this means replacing some function of π₯ with π’. So, we also replace dπ₯ in terms of dπ’ and then we integrate in terms of π’. We can then replace π’ with our function of π₯ to obtain the answer. So, letβs say π’ equals negative 12π₯ add 26. Then, we know in our integral, we can replace negative 12π₯ add 26 with π’. But weβre also going to need to replace dπ₯ with something in terms of dπ’.

We do this by first of all finding dπ’ by dπ₯. This is the derivative of π’ with respect to π₯. As π’ is equal to negative 12π₯ add 26, when we differentiate this with respect to π₯, we just get negative 12. And although dπ’ by dπ₯ is not a fraction, we do treat it a little bit like a fraction when weβre trying to manipulate it in this way. So, by rearrangement, we find that dπ₯ is equal to negative one over 12dπ’.

So now, we have everything we need to make the substitution. Our integral was the integral of negative 12π₯ add 26, which weβre replacing with π’, multiplied by the square root of negative 12π₯ add 26. So, this is multiplied by the square root of π’. And then, we know that we can replace dπ₯ with negative one over 12dπ’. Now, letβs simplify our integral a little bit.

Firstly, we know that we can bring constants in the integral outside to the front of the integral. So, we can bring the negative one over 12 out to the front of the integral. Another thing to notice is that we can replace the square root of π’ using the fact that the square root of π’ is equivalent to π’ to the power of one-half. Now, from here, there is a rule that we can apply thatβs going to simplify π’ multiplied by π’ to the one-half power. And thatβs the rule that says that π’ to the π power multiplied by π’ to the π power is equal to π’ to the power of π add π. And as π’ on its own is equivalent to π’ to the power of one, this is π’ to the first power add π’ to the half power, and thatβs just π’ to the one add one-half power. And as one add one-half is three over two, this is just π’ to the three over two power.

So now weβve simplified our integral, letβs integrate. We recall the power rule of integration, which tells us that to integrate π’ to the power of three over two, we add one to the power and then divide through by this new power. Adding one to the power gives us the new power of five over two, and then we need to divide by five over two. But dividing by five over two is just the same as multiplying by two over five. So this gives us negative one over two two-fifths π’ to the five over two power. And because we donβt have any limits of integration, we need to remember to add a constant, so plus πΆ. Letβs now simplify this by multiplying together the negative one over 12 with the two over five. This gives us negative two over 60π’ to the five over two power plus πΆ. But we can actually cancel down the two over 60 to one over 30.

So now weβve integrated our function and weβve simplified it, thereβs one more thing we need to remember to do. Because we said that π’ is equal to negative 12π₯ add 26, we need to remember to replace π’ with negative 12π₯ add 26 for our final answer. Doing so gives us the answer of negative one over 30 multiplied by negative 12π₯ add 26 to the five over two power plus πΆ.