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

7–64. Integration review Evaluate the following integrals.

28. ∫ (3x + 1) / √(4 - x²) dx

Accepted Answer

Welcome back, everyone. Evaluate the integral, integral of 5 X minus 2, all divided by square root of 9 minus X squared DX. So essentially what we're going to do for this problem is rewrite the integral as the difference between two integrals and factor out the constants. So for the first integral we can factor out 5. We have X in the numerator. Our denominator is the radical term, so we're going to introduce square root of 9 minus X 2 D X. And now we are subtracting 2 rights so we can factor out 2 integral. D X divided by square root of 9 minus x2. So we have a difference of two integrals. Let's consider each integral separately. Starting with the first one, we want to integrate X divided by square root of 9 minus X 2 D X, and for this integral we're going to introduce. Use substitution. Let's suppose that U is equal to 9 minus X2. So that the derivative DU divided by DX is -2 X according to the power rule. And then we can rearrange this relationship to show that our numerator XDX. Is equal to -1/U. Meaning our integral becomes integral of XDX, so that's -12 DU. We can factor out -12 and we end up with DU divided by square root of u. And now this is a basic integral we can evaluate it using the power rule. We get -12 integral of U to the power of -12 DU. According to the power rule, we get -12 multiplied by U to the power of 1/2 because we're adding 1 to the original exponent and dividing by 1/2, right? We have to divide by the new exponent. And then we add a constant of integration C1. It's C1 because we have two indefinite integrals, so we're going to differentiate between the two constants of integration. Now -12 divided by 1/2 is simply -1, so we get minus. And then use the power of one half is square root of you. Plus C1. Substituting back for you, we get negative square root off. 9 minus X2 plus C1. So that's our first integral. Later on, we will multiply by 5, but for now, we can just label this as our first answer that we will use later on. But now what we can do is simply focus on the 2nd integral. In the role of TX. Divided by square root of 9 minus X2. And now this integral has a form of integral. Of the X. Divided by a square root of A 2 minus X2. Where a is a constant, right? And our A is 3 because 3 squared is 9. Now, the value of this integral is inverse of sine. Of X divided by A plus C in a general case, right? So, if we use this formula, we can essentially conclude that the integral of D X divided by square root of 9 minus X2. Is inverse of sign. Of X divided by A and A is square root of 9, which is 3, right? So inverse of sine of X divided by 3 plus our constant of integration C2. And that's our second answer that we're going to use. And then we can use these answers to evaluate the final result. So, 5 multiplied by the value of the first integral would be -5. Square root of 9 minus X2d we can ignore the constant for now. And then we're subtracting 2 multiplied by the result. For the 2nd integrals we're subtracting 2. In verse of sign. Of X divided by 3 and then we can combine our constants. We can just write plus C, right? C 1 + C2 would be equal to C. Of course they are multiplied by 5 and 2 respectively, but Essentially, whichever form of the sum of those constants we get, we can just write it as plus C. and that would be our final answer for this problem. Thank you for watching.

7–64. Integration review Evaluate the following integrals.
28. ∫ (3x + 1) / √(4 - x²) dx

Key Concepts

Integration Techniques

Trigonometric Substitution

Definite vs. Indefinite Integrals

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