63–68. Definite integrals Evaluate the following definite inte...

Question

63–68. Definite integrals Evaluate the following definite integrals. Use Theorem 7.7 to express your answer in terms of logarithms.

∫₋₂² dt/(t² – 9)

Accepted Answer

Compute the integral from -3 to 2 of DT divided by T2 minus 25 and express your answer in terms of logarithms. Now, to solve this, we're going to rewrite our equation. We know this first of all, that our denominator can be factored. So we have the integral from -3 to 2. DT divided by T + 5, T minus 5. Let's go ahead and modify this. We have 1 divided by T + 5 multiplied by T minus 5, which by using partial fractions can turn into a divided by T + 5, plus B divided by T minus 5. Let's go ahead And solve the system We'll multiply both sides by the denominator, T + 5, T minus 5, to get 1 equals A multiplied by T minus 5, plus B multiplied by T + 5. Let's go ahead and solve this. Let's let T equals 5 1st. We get 1 equals A multiplied by 5 minus 5, plus B multiplied by 5 + 5. A will cancel, giving us 1 equals 10B, or B equals 10. Let's let T equals -5 next. 1 equals a multiplied by -5 minus 5. Plus B multiplied by 5 + 5. We get 1 equals -10A. And if we divide both sides by negative 10. A equals -10. We now have our partial fractions. We can rewrite this as -10 multiplied. By 1 divided by T + 5. Plus 11 multiplied by 1 divided by t minus 5. We want the antiderivative of this now. From 3 to 2, DT. We can solve this by factoring out 1/1 1st to get 1/1 multiplied by the integral from -3 to 2. Of negative 1 divided by T + 5. Plus 1 divided by T minus 5. DT And finally, I'm just going to rewrite this to have the positive term first. 10th multiplied by the integral from -3 to 2, of 1 divided by T minus 5 minus 1 divided by T + 5 DT. Now let's take our antiderivative. We will get 1/1 multiplied by the natural log of T minus 5. Minus the natural log of T + 5. From -3 to 2. From here we can plug in our values. We get 1/10 multiplied. By the natural luck And we first plug in 2. We get to minus 5. Minus Sinatra log of 2 + 5. Then we plug in Our -3 Natural light of -3 minus 5. Minus natural log of -3 + 5. Now we can take the absolute value of our negatives to get positive terms. This will simplify. We get the natural log of 3 minus the natural log of 7. Minus the natural log of 8 plus the natural log of 2 when distributing are negative. Now by log rules, we can rewrite our values. We will see We had the natural log of 3 plus the natural log of 2. Minus the natural log of 7 plus the natural log of 8s. Which can then be rewritten. As one tins multiplied by the natural log. Of 3, multiplied by 2, divided by the natural log. Of 7 multiplied by 8. Or in other words we can rewrite this once more. To be under the same. Luck Let's go ahead and simplify this now. We get 1/10 natural log of 6 divided by 56, which gives us 1/11 natural log of 3 divided by 28. This would be the final answer to our problem. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

63–68. Definite integrals Evaluate the following definite integrals. Use Theorem 7.7 to express your answer in terms of logarithms.∫₋₂² dt/(t² – 9)

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63–68. Definite integrals Evaluate the following definite integrals. Use Theorem 7.7 to express your answer in terms of logarithms.
∫₋₂² dt/(t² – 9)