23-64. Integration Evaluate the following integrals.
26....

Question

23-64. Integration Evaluate the following integrals.

26. ∫₀¹ [1 / (t² - 9)] dt

Accepted Answer

Welcome back, everyone. Compute the integral from 2 to 3 of 1 divided by Y2 minus 1DY. For this problem we're going to use partial fraction decomposition. Let's notice that we can rewrite 1 divided by Y2 minus 1 S 1 divided by. Y minus 1 multiplied by Y + 1, right, because we have a difference of squares in the denominator. So now what we're going to do is simply rewrite this expression as a divided by our first linear factor Y minus 1 plus B divided by our next linear factor Y + 1. Our goal is to identify the coefficients A and B. So what we can do is basically go backwards and find the common denominator. We already know that it is Y minus 1 multiplied by Y + 1. And for that purpose, A is multiplied by Y + 1, so we get A multiplied by Y + 1 in the numerator, plus B multiplied by Y minus 1, right? And then we can expand our numerator, right? So we're going to rewrite our denominator. And now expanding the numerator we get A Y + A + B Y minus B. Now let's go ahead and factor out why we get. A plus B multiplied by Y and of the remaining terms are A minus B. We're dividing that by y minus 1 multiplied by Y plus 1. If we look at the original expression, our numerator contains 1 in the numerator, right? So A plus B must be equal to 0. Because we do not have any white terms in the numerator. And then A minus B must be equal to 1 because A minus B represents a coefficient, right, which is a 1 in our case. So, A minus B is equal to 1. We want to solve this equation. From the first equation, we are going to show that A is equal to negative B. And then substituting negative B into the second equation for A, we get negative B minus B, which is -2 B equals 1. From here we can show that B is equal to -12 and then A, which is negative B, is going to be 1/2. So we can now show that we're integrating. A divided by Y minus 1, which is 1/2. Divided by y minus 1. Plus B, which is 1/2, we can just write minus 12 divided by Y + 1. The why we're going to add brand Cs, and we're going to show that we're integrating from 2 to 3. Let's go ahead and factor out 1/2. We get 1/2. Integral from 2 to 3 of. 1 divided by Y minus 1 minus 1 divided by Y plus 1 D Y. And then we have a basic integral, right, one divided by Y, the Y, in terms of an integral is going to be a n of the absolute value of Y, right? If we have a linear term such as Y minus 1, we just have to divide by the slope of the linear term. So in this case, our slope is 1, meaning nothing changes according to the reverse chain rule. We basically get our expression for the integral as follows. First of all, we're going to add 1/2 because we have factored out 1/2. Imparencies we're going to have a n of the absolute value of Y minus 1 as the integral of 1 divided by Y minus 1 minus Ln of the absolute value of Y + 1. And we want to evaluate from 2 to 3. Using the properties of logarithms we can simplify slightly and show that this is 1/2. Multiplied by Ln of the absolute value of Y minus 1. Divided by Y + 1. And we want to evaluate this expression from 2 to 3, so let's go ahead and do that. We got 1/2. In pricy, we take a land of the absolute value of 3 minus 1 gives us 2. Divided by 3 + 1 is equal to 4, right? So we get LN of 2 divided by 4, which is LN of 1/2. We also don't need to use the absolute value because this is a positive number. And we're going to subtract LN of the absolute value of 2 minus 1 is equal to 1 divided by 2 + 1, which is 3. Again, 1/3 is greater than 0. So we're going to add renies instead of the absolute value. And now let's simplify. This is going to be 1/2 multiplied by LN of using the properties of logarithms, we can divide 1/2 by 1/3, so we get 3 halves, right? Well, once we have our final answer, that's 1/2 Len of 3 halves. Thank you for watching.

23-64. Integration Evaluate the following integrals.
26. ∫₀¹ [1 / (t² - 9)] dt

Key Concepts

Improper Integrals

Partial Fraction Decomposition

Definite Integration Techniques

Watch next