Evaluating integrals Evaluate the following integrals. &n...

Question

Evaluating integrals Evaluate the following integrals.

∫ 𝓍⁷ √(𝓍⁴ + 1d𝓍)

Accepted Answer

Hello. In this video, we are going to be evaluating the following integral. We are given the integral of X 9th multiplied by the square root of X 5 plus 3. Notice that we are given a composite function inside of the integral. We have the square root of X 5 + 3, which is a function of X inside another function of X. This is an indication that we should approach this integral by using a use substitution. We are going to allow a term U to equal to the innermost function of the composite function, and that is going to be X 5th power plus 3. Now, since we are transforming this function from terms of X to terms of U, we have to come up with a substitution for DX in terms of you, and for every X term in terms of you. So, in order to find a substitution for DX in terms of you, we will need to take the derivative of both sides of our U substitution. By taking the derivative of U with respect to X, we will end up with DU is equal to 5x the 4th power, DX. And if we were to divide both sides of the substitution by 5, we will end up with 1/5 DU is equal to x 4th DX. So currently we are working with two substitutions. We are substituting in the innermost function with U, but our 150 U substitution is acting as a substitution for our DX term, but only for multiples of X. Currently, we have 9 extra multiples of X. So in order for us to utilize this substitution in multiples of 4, or for X to the 4th power, we will need to rewrite this X the 9th power. By separating this as a product of X, we can rewrite X the 9th as X 5th power, multiplied by X the 4th power, and this will be multiplied by the square root of X 5 + 3 DX. Now, this substitution for DX will allow us to substitute DX and X the 4th in terms of U, and our original U substitution is now substituting X 5th plus 3 in terms of you. But what about this extra X to the 5th? Well, we can solve for X to the 5th in our original U substitution. And that will give us x 5 is equal to U minus 3, and this is going to be a substitution for x 5th in terms of you. So by utilizing the three substitutions, we can rewrite the integral in terms of you to be the quantity U minus 3 multiplied by 1/5 DU. Multiplied by the square root of you which can also be rewritten as you to the half power. OK. Now let's just go ahead and reorder some of the terms in our integral. We can pull out the 15th as a constant in front of the integral, and this will allow us to rewrite the integral to be U minus 3, multiplied by U to the half power DU. And now we can use algebraic methods to solve for this integral. So what we will go ahead and do is distribute you to the half power into U minus 3. That will allow us to rewrite the integral as 1/5th multiplied by the integral of U 32 minus 3 multiplied by U D U. And now we will go ahead and take the antiderivative of each of the terms of you. That is going to leave us with 15th multiplied by you to the 3 halves plus 1. Divided by 3 halves plus 1. -3. -3, multiplied by the power plus 1, divided by 1/2 + 1. And because this is an undefined integral, we'll be adding a constant term C at the end of this. Now by simplifying algebraically, this is going to simplify to be 1/5 multiplied by the quantity. 2/5 Multiplied by you to the 5 halves power, minus. 2 Multiplied by U to the three halves power plus C. Notice that we can factor a 2. From the groups of you. So if we were to factor out a 2, we will end up with 25ths multiplied by 1/5th multiplied by you. To the 5 has power. Minus U to the three halves power plus C. And finally, all we need to do is we need to resubstitute you in terms of X. Recall earlier that our U substitution was X 5 + 3. So in our antiderivative, we are going to substitute every term of U with X 5 + 3, and that is going to leave us with 25ths multiplied by 1/5, multiplied by X the 5th + 3. That quantity rates the 5 halves power. Minus the quantity x 5 + 3. Raise to the three halves power plus C, and this is going to be the simplest form of the solution to the integral. And with that being said, the overall solution to this problem is going to be a. So I hope this video helps you in understanding how to approach this type of integral, and we will go ahead and see you all in the next video.

Evaluating integrals Evaluate the following integrals.

∫ 𝓍⁷ √(𝓍⁴ + 1d𝓍)

Key Concepts

Integration

Substitution Method

Definite vs. Indefinite Integrals

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