7–64. Integration review Evaluate the following integrals.
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Question

7–64. Integration review Evaluate the following integrals.

51. ∫ from -1 to 0 of x / (x² + 2x + 2) dx

Accepted Answer

Welcome back, everyone. In this problem, we want to evaluate the integral of X divided by X2 + 4 X + 5 with respect to X between the limits of 1 and 2. Now to evaluate this integral, it would be a good idea for us to substitute. But before we do that, recognize that if in our denominator, OK, notice that our denominator X2 + 4 X + 5 can be written as X2 + 4 X + 4 + 1. Now why would we do that? Well, notice here that X2 + 4 X + 4 is a perfect square, which means we can rewrite it as X + 2 are squared. Now that would be pretty helpful. Because if we let you be equal to X + 2, then that means we can take our denominator, X2 + 4 X + 5, OK, and rewrite it as U2 + 1. But notes that we have a way to write our denominator in terms of you. Let's see if we can write the rest of terms uh related to you or in terms of you, right? Now for our numerator, if you equals x + 2, OK, then this would imply that x would be equal to U minus 2. And if X equals u minus 2, then the derivative of U with respect to X would be equal to 1, which implies that DU is going to be equal to D X. Next, let's start thinking about our limits. Now we already know that for X, the limits are between 1 and 2. So for our lower limit, when X equals 1. Then you is gonna be equal to 1 + 2 or 3. And when X equals 2 U equals 2 + 2 or 4. So now, if we summarize everything we have, our denominator will now be U2 + 1. Our numerator U minus 2, DU will be equal to DX and our limits will be between 3 and 4. Now that we have all this information we can substitute for you. No substituting for you, OK? Then we can rewrite the derivative of, sorry, the integral, my apologies, between the limits of 1 and 2 of X divided by X + 2 or squared + 1 with respect to X. As the integral between 3 and 4 of U minus 2 divided by U2 + 1 D U. And we can split up our numerator even further to no get. Between the limits of you divided by U2 + 1 with respect to U minus 2 multiplied by the integral of 1 divided by U2 + 1 with respect to you. Now let's go ahead and find the integral of each term. Here we are now evaluating each term. I know to do that, let's start with our first term. OK, let me just put a line here to separate all this work that uh we're gonna be doing. So starting with our first term, OK, we're gonna be finding the integral between the limits of 3 and 4 of you divided by U2 + 1 with respect to you. Again we can solve by substituting. Let's set the to be equal to U 2 + 1. Then that means. The derivative of V with respect to you is going to be equal to 2, OK, which then implies that DV will be equal to 2 with DU. And thus UDU is gonna be equal to a half of DV and we want UDU because if you notice, UDU is in our integral. Know that we have it, if we also change the limits, OK. When you equals 3, then V equals 32 + 1, which is 10, and when you equals 4, V is going to be 17. So thus we can rewrite our integral then as the integral, or sorry, a half, let's take out our constant, a half of the integral of 1 divided by the DV between the limits of 10 and 17. The integral of 1 divided by V with respect to V is going to be equal to the natural log of V. So we can rewrite this then as half of the natural log of V between the limits of 17 and 10. And when we substitute those limits, when we put those limits into our expression. OK, then we're gonna get a half of the natural log of 17 minus the natural log of 10. And by the laws of logs, we could rewrite that as the natural log of 17/10, OK? So the integral of Udivided by U2 + 1 with respect to you is a half of the natural log of 17/10. Now let's go ahead and see if we can integrate our second term. Now in our second term, we're finding. The negative 2 multiplied by the integral between the limits of 3 and 4 of 1 divided by U2 + 1 U. Now this one might be a little bit more straightforward because if we can recall that the integral of 1 divided by U quad plus 1 DU is going to be equal to the arc tangent of U plus a constant. So in that case then that means this is going to be equal to -2 multiplied by the arc tangent of U between the limits of 3 and 4. And if we substitute those limits, then we're gonna get negative 2 multiplied by the arc tangent of 4 minus the arc tangent of 3. Now that we have our results, our integrals for two expressions, we can combine the results, OK. And now by combining the results. Then that means our final integral is going to be a half of the natural log of 17/10. Minus 2 multiplied by the arc tangent of 4 minus the arc tangent of 3. This is the integral of X divided by X2 + 4 x + 5 DX between the limits of 1 and 2. Thanks a lot for watching everyone. I hope this video helped.

7–64. Integration review Evaluate the following integrals.
51. ∫ from -1 to 0 of x / (x² + 2x + 2) dx

Key Concepts

Definite Integrals

Integration Techniques

Fundamental Theorem of Calculus

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