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

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

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Determine ∫ 8𝑥(8𝑥 + 9)² d𝑥 by using the substitution method.

02:28

### Video Transcript

Determine the integral of eight 𝑥 times eight 𝑥 plus nine squared with respect to 𝑥 by using the substitution method.

In this example, we’ve been very explicitly told to use the substitution method to evaluate this integral. Usually, we will look to choose our substitution 𝑢 to be some factor of the integrand whose differential also occurs, albeit some scalar multiple of it. Here though, it’s not instantly obvious what that might be. Instead, then, we try choosing 𝑢 to be some more complicated part of the function, perhaps the inner function in a composite function.

Let’s try 𝑢 equals eight 𝑥 plus nine. This means that d𝑢 by d𝑥 is equal to eight. And we can treat d𝑢 and d𝑥 as differentials. Remember, d𝑢 by d𝑥 is not a fraction but we certainly treat it like one when performing integration by substitution. We can say that d𝑢 is equal to eight d𝑥. Or, equivalently, an eighth d𝑢 is equal to d𝑥. Now, this isn’t instantly helpful. As if we replace d𝑥 with an eighth d𝑢 and eight 𝑥 plus nine with 𝑢, we’ll still have part of our function, that’s the eight 𝑥, which is in terms of 𝑥.

But if we look back to our substitution, we see that we can rearrange this. We subtract nine from both sides, and we see that eight 𝑥 is equal to 𝑢 minus nine. Then, our integral becomes 𝑢 minus nine times 𝑢 squared multiplied by an eighth d𝑢. Let’s take this eighth outside of the integral and then distribute the parentheses, and we see we have a simple polynomial that we can integrate.

The integral of 𝑢 cubed is 𝑢 to the fourth power divided by four. The integral of negative nine 𝑢 squared is negative nine 𝑢 cubed divided by three. And we mustn’t forget 𝐶, our constant of integration. We can simplify nine 𝑢 cubed divided by three to three 𝑢 cubed. But we mustn’t forget to replace 𝑢 with eight 𝑥 plus nine in our final step.

When we do, we see that our integral is equal to an eighth times eight 𝑥 plus nine to the fourth power divided by four minus three times eight 𝑥 plus nine cubed plus 𝐶. When we distribute our parentheses, we see we have our solution. It’s one over 32 times eight 𝑥 plus nine to the fourth power minus three-eighths times eight 𝑥 plus nine cubed plus 𝐶.

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