2–74. Integration techniques Use the methods introduced in Sec...

Question

2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.

18. ∫ (from 0 to √2) (x + 1)/(3x² + 6) dx

Accepted Answer

Hello. In this video, we are going to be evaluating the definite integral using the appropriate integration techniques. So we are given the integral from 0 to 1 of X + 4 divided by X2 + 5. Now, if we were to apply a U substitution in the denominator, we wouldn't get a U substitution that would help us eliminate anything in the numerator. So instead, because we have a sum of two terms in the numerator, let's go ahead and break up this integral into a sum of integrals. So we are going to start by splitting up the integrand into a sum of fractions. That is going to give us the integral from 0 to 1 of X divided by X2 + 5, plus 4 divided by X2 + 5DX. Then if we assign each of the terms their own integration symbol, we get the sum of the integral from 0 to 1 of X divided by X2 + 5, plus the integral from 0 to 1 of 4 divided by X2 + 5 D X. What we are going to do is we are going to solve each of the integral integrals independently. Starting with the first integral, we have the integral from 0 to 1 of X divided by X2 + 5 DX. Let's go ahead and apply a U substitution to the denominator and allow U to equal to X2 + 5. Then, when we find a substitution for DX by integrating the U substitution with respect to X, we get that DU is equal to 2 XDX. Now, we have a DX. And XDX in the in the integral, but we need to move over that coefficient of 2. So that will give us 1/2 DU is equal to XDX. And then once we applied the use substitution. We will be left with the integral. From A to B of 1 divided by U, multiplied by 1/2 DU, but we will move the 12 to the front of the integral as a constant. This will simplify to be one half multiplied by the natural logarithm of U evaluated from A to B, and once we backs substitute our U substitution in terms of X, we will be left with 1/2 multiplied by the natural logarithm of X2 + 5, and this will be evaluated from 0 to 1. We will go ahead and set this value to the side for now. Now what we are going to do is we are going to solve for the 2nd integral. Now, the second integral is the integral from 0 to 1 of 4 divided by X2 + 5. We will go ahead and move the 4 to the front of the integral, leaving us with 4 multiplied by the integral from 0 to 1 of 1 divided by X2 + 5 DX. Now, we can solve this integral by using the integration of the tangent inverse function, that is given as the following. If we have an integral of the form, X2 plus A2, this is going to simplify to be 1 divided by A, multiplied by tangent inverse of X divided by A. Here, our A square term is equal to 5, and that means our A term is going to be the square root of 5. So by using this identity, the value of the integral is going to be 4, multiplied by 1 divided by the square root of 5. Multiplied by the tangent inverse of x divided by the square root of 5. And this will be evaluated from 0 to 1, and by rationalizing the denominators and multiplying by 4, this will simplify to be 4 square root of 5 divided by 5, multiplied by tangent inverse of 5 of X multiplied by square root of 5. Divided by 5, and this will be evaluated from 0 to 1. What we have here is the values of both of the integrals, and they are both being evaluated from 0 to 1. So, if we plug this back into the original sum that we were working with, this is going to give us the sum of 1/2 multiplied by the natural logarithm of X2 + 5 + 4 multiplied by square root of 5 divided by 5, multiplied by the tangent inverse. Of X multiplied by square root of 5 divided by 5, and this entire thing will be evaluated from 0 to 1. Now what we're going to do is we're just going to plug in the values of the boundaries. So by plugging in 1, we will be left with one half multiplied by the natural logarithm of 6 plus 4 square root of 5 divided by 5, multiplied by the tangent inverse of square root 5 divided by 5. Then we will plug in 0, and that is going to give us negative quantity 1/2 multiplied by the natural logarithm of 5, plus, Plugging in 0 into the tangent inverse is going to give us tangent inverse of 0, and that entire thing is going to evaluate to be zero. So we will add 0. Now we just need to simplify. The only thing we can do here is distribute the negative sign into the natural logarithm, giving us negative 12 multiplied by the natural logarithm of 5, and by using our natural logarithm properties, we will be left with 1/2 multiplied by the natural logarithm of 6 divided by 5, plus 4 square root 5 divided by 5, multiplied by the tangent inverse of square root 5 divided by 5. And this is going to be the value of the given definite integral. And with that being said, the overall solution to this problem is going to be B. So, I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.18. ∫ (from 0 to √2) (x + 1)/(3x² + 6) dx

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2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.
18. ∫ (from 0 to √2) (x + 1)/(3x² + 6) dx