Video Transcript
Determine the integral of 𝑥 multiplied by two 𝑥 squared minus nine all cubed.
Okay, to actually solve this problem and integrate this expression, what we’re actually gonna use is we’re gonna use substitution. This method is also known as the reverse chain rule. So the first stage to actually solve by substitution is to decide which part is going to be 𝑢. So I’ve said that 𝑢 is equal to two 𝑥 squared minus nine. And now I’m actually gonna differentiate this because I want to find 𝑑𝑢 𝑑𝑥.
If I differentiate two 𝑥 squared minus nine, I get four 𝑥. And just to quickly remind us how we got that, well, actually, you multiply the exponent by the coefficient. So two multiplied by two gives us four. And then you subtract one from the exponent. So instead of two, we have one. So we’ve got four 𝑥. And the reason we actually do this is because we’re looking to find 𝑑𝑥. So what we’re gonna do now is rearrange to actually find 𝑑𝑥.
So first of all, we actually multiply by 𝑑𝑥. So we’re gonna get 𝑑𝑢 is equal to four 𝑥 𝑑𝑥. And then we actually divide both sides by four 𝑥 to get 𝑑𝑢 over four 𝑥 is equal to 𝑑𝑥. So now we actually have chosen what 𝑢 is and we found 𝑑𝑥. What we’re gonna do is actually substitute these back into our original expression. And when we do that, we now have it in the form that we want the integral of 𝑥 multiplied by 𝑢 cubed multiplied by 𝑑𝑢 over four 𝑥.
Well, this still leaves us with a little bit of a problem because the reason we used substitution is actually so that we can create an expression that’s easier to integrate. However, we still got 𝑥 and 𝑢 in our expression. But actually, if we have another inspection of the expression we had, we can say that if we divide through by 𝑥, we can actually eliminate our 𝑥 terms. And this leaves us with a much more straightforward integration because we’ve now got to integrate 𝑢 cubed over four.
So this gives us 𝑢 to the power of four over 16 plus 𝑐 cause don’t forget the constant of integration. And just reminds us how we did that, we raised the exponent by one, so went from 𝑢 cubed to 𝑢 to the power of four. And then we divided by the new exponent. But because we already had one over four or 𝑢 cubed over four, we then multiply four by the new exponent, which is four, to give us 16 as our denominator.
Okay, great! So now what’s the next step? What do we do? So the final thing that we need to do is actually substitute back in 𝑢 is equal to two 𝑥 squared minus nine. So therefore, when we do, we can say that the integral of 𝑥 multiplied by two 𝑥 squared minus nine all cubed is equal to one over 16 two 𝑥 squared minus nine all cubed plus 𝑐.
