2–74. Integration techniques Use the methods introduced in Sec...

Question

2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.

68. ∫ (from -1 to 1) dx/(x² + 2x + 5)

Accepted Answer

Welcome back, everyone. In this problem, we want to compute the definite integral of DX divided by X grad + 4 X + 10 on the interval from 0 to 2. Now, before we compute this integral, it would be helpful for us to rewrite our denominator to include the perfect square. Because note here that we could rewrite X2 + 4 X + 10 as X2 + 4 X + 4 + 6. Now, X2 + 4 X + 4 is a perfect square. So we could rewrite this as X + 2 squad + 6. Now why would we do that? Well, in this way, it makes it easier for us to substitute because now if we let you be equal to x + 2, then our denominator could be rewritten as U2 + 6. Let's try to include substitution in the rest of the problem. Now if you equals x + 2, then the derivative of you. With respect to X is going to be equal to 1, which implies that DU is equal to DX. And know when thinking of our limits when X equals 0, then you will be equal to 0 + 2 or 2. And when X equals 2, then U equals 2 + 2 or 4. So now we have enough information to substitute you into our integra. And now by substituting, OK, then our integral should now become the integral between the limits of 2 and 4 of DU divided by U2 + 6. Now what do we know that can help us to solve in our problem? Well, recall in our standard form for integrals that the integral of DU divided by U2 plus a 2 or the sum of two squared terms is gonna be equal to 1 divided by a multiplied by the arc tangent of U divided by A. OK, where in our problem A a squared would be 6, so A is going to be equal to root 6. So no, that tells us for our problem, then all of this is going to become one divided by root 6 multiplied by the arc tangent. Of you divided by A or which is root, sorry, A, which is route 6, OK. Between the limits of 2 and 4, OK? So now in this case, this would be the integral of our expression. No, we can go ahead and substitute our limits. Because no, this means we're really saying it's gonna be 1 divided by route 6 of the arc tangent of 4 divided by route 6, OK. Minus the arc tangent of 2 divided by route 6. And now, in this case, OK, if we rationalize by multiplying through by route 6, then we're gonna have root 6 divided by 6 multiplied by the arc tangent of 4 route 6 divided by 6 minus the arc tangent of 2 route 6 divided by 6. I know we can simplify that even further. In the arguments for our tangent, because 4 route 6 divided by 6 can be rewritten as 2 routes 6 divided by 3. And 2 Route 6 divided by 6 can be rewritten as a third of Route 6. Therefore, our final answer is root 6 divided by 6 multiplied by the arc tangent of 2/3 of route 6 minus the arc tangent of 1/3 of route 6. Thanks a lot for watching everyone. I hope this video helped.

2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.
68. ∫ (from -1 to 1) dx/(x² + 2x + 5)

Key Concepts

Integration Techniques

Definite Integrals

Completing the Square

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