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

7–64. Integration review Evaluate the following integrals.

49. ∫ √(9 + √(t + 1)) dt

Accepted Answer

Welcome back, everyone. Evaluate the integral. Integral of square root of 16 plus square root of X + 2 DX. For this problem we're going to use a U substitution. Let's suppose that U equals 16 plus square root of X + 2. And as we understand, our next step is to identify the derivative of U with respect to X. The derivative of 16 is 0, the derivative of square root is 1 divided by 2 square root of our function, so that's X + 2. According to the chain rule, we're multiplying by the derivative of X plus 2, which is just 1, right? So now what we're going to do from here is express DX. And we can show that the X is DU multiplied by 2 square roots of X + 2. Well done. Now notice that square root of X + 2 from our U substitution can be expressed as U minus 16. We can subtract 16 from both sides, right? Which means that DX is equal to 2 multiplied by U minus 16 DU. So now we have our complete expression for DX in terms of U and DU. Well done. So now going back to the original integral, we can show that the integral becomes square root of. First of all, we have our you. And now we can replace the X with. 2 multiplied by U minus 16 DU. Let's factor out the constant. This is going to be 2 integral. And then we have square root of U, which is simply U to the power of 1/2 multiplied by U minus 16, multiplied by DU. Now let's distribute. This gives us 2 integral to the power of three halves. Minus 16. Use the power of one half. The EU And now what we can do is simply split it into two integrals, right? So essentially this is going to be 2 integral of U to the power of 3 halves the minus 2 multiplied by 16 is 32 integral to the power of 12 DU. And now let's integrate, so we got 2 multiplied by. You to the power of 5 halves according to the power rule, we are adding 1 to the original exponent and dividing by the new exponent. So if we're dividing by 5 halves, this means we're multiplying 2 by 2 and dividing by 5. So that'd be 4/5s, we can just write 4 divided by 5 in front. So 4/5 uses the power of 5 halves. That's our first part. And then we are subtracting 32 U to the power of three halves, because 1/2 plus 1 is 3 halves divided by 3 halves. So we're multiplying 32 by 2 and dividing by 3, so that's 64 divided by 3. And then we add a constant of integration C. So what we're going to do from here is simply perform factorization. We can factor out 4. Divided by 15, because the common denominator of 5 and 3 is 15. Which leaves us with, of course, another term we have to consider use the power of 5 halves and use the power of three halves. We can factor out U to the power of three halves, and then imprecies. We're going to have 3 you considering the first term minus. Haiti Plus a constant of integration C. So now we have a factored form and we can substitute. You with an expression in terms of X. So we get 4 divided by 15. Now let's recall that use the power of three halves is simply square root of you cubed. So we have square root. Off You ube, so that'd be 16. Plus square root of X + 2. We are essentially cubing this whole expression. To make it simpler, we can just express it in terms of the whole exponent, right? So instead of using a radical term, we can just say 16 + square root of X + 2 is raised to the power of three halves, right? And then considering our next factor, we have 3 Us, so that's 3 multiplied by 16 plus square root of X + 2. 80 And finally, plus C. So now 3 multiplied by 16 is 48 and 48 minus 80. Is equal to. 32 so we can just distribute and say that we're subtracting 32. And this leaves us with 3 square root of X + 2 when we. Finish our distribution, right? So the, so our final answer would be. 4 divided by 15 multiplied by 16 plus square root of X + 2 raises the power of 3 halves. Multiplied by 3 square root of X plus 2 minus 32 + C. And that will be our final answer for this problem. Thank you for watching.

7–64. Integration review Evaluate the following integrals.
49. ∫ √(9 + √(t + 1)) dt

Key Concepts

Integration Techniques

Substitution Method

Definite vs. Indefinite Integrals

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