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.

15. ∫ (from 1 to 2) (3x⁵ + 48x³ + 3x² + 16)/(x³ + 16x) dx

Accepted Answer

Hello. In this video we are going to be evaluating the following integral. We are given the integral from 4 to 6 of the quantity 4 x 5th power plus 36 x cubed plus 6 X2 + 27 divided by the quantity X cubed plus 9 X. Now, when we are looking at this integral, notice that the overall order of the numerator is higher than the order of the denominator. What this means is that we can simplify the integran by doing polynomial long division. So, let's go ahead and set up the polynomial long division. We have X cubed plus 9 X, and that is going to divide the numerator, 4 X to the 5th power, plus 36 X cubed, plus 6 X squad, plus 27. Now, in order to perform this polynomial long division, we need to try to match this first term of the of the denominator with the first term of the numerator. So, we can multiply the denominator by 4 X squared, and that is going to give us the quantity 4 X to the 5th power plus 36 X cubed. Then we need to subtract this quantity. This is going to eliminate the X to the 5th and X cubed term, leaving us with the remainder of 6 X2 plus 27. So, what this means is that we can now rewrite the integral in a simplified form by the polynomial long division, and that is going to give us the integral from 4 to 6 of 4 X squad, plus the remainder 6X2 + 27 divided by X cubed plus 9 XDX. Now we can further simplify the second term in the integran by using partial fraction decomposition. If we were to factor the denominator, we can rewrite this part to be 6 X2 + 27 divided by X multiplied by X + 9, X2 + 9. And the partial fractions of the denominator are going to simplify to be a divided by X plus B X + C divided by X2 + 9. Now, if we were to multiply by the lowest common denominator, we get 6 X2 plus 27 equal to A multiplied by X2 + 9, plus the quantity B X plus C multiplied by X. And if we were to distribute and reorder the terms, we will be left with 6 X2 + 27 is equal to A plus B multiplied by X2 + C X plus 9 A. Now, what we are going to do is we are going to create a system of equations based off of this equation. By matching the coefficients of the X2 terms, the X terms, and the constants, we will be left with a plus B is equal to 6, C is equal to 0, and 9A is equal to 27. Now, this is already one of the values we need for the partial fraction decomposition. From this last term here, we get that A is equal to 3, and by plugging in this value of A, we get that B is equal to 3. And so if we were to plug this back into the partial fraction decomposition, we can further rewrite the integral to be the integral from 4 to 6 of 4 X squad, plus 3 divided by X plus 3 X divided by X2 + 9 DX. Now what we are going to do is we're going to split this up into a sum of definite integrals. So, we are going to give each term its independent integral symbol from 4 to 6. Then what we are going to do is we're going to solve for each of these independently. Starting with the first integral, we have the integral from 4 to 6 of 4 X squared. If we take the antiderivative, we will be left with 4/3 X cubed, and this will be evaluated from 4 to 6. We will go ahead and set this to the side for now. For the second integral, we have 3, the integral of 3 divided by X. If we move the 3 in front of the integral as a constant, we get 3 multiplied by the integral from 4 to 6 of 1 divided by X, and this is going to simplify to be 3 multiplied by the natural logarithm of X, and this will be evaluated from 4 to 6. We will come back to this later. Now we just need to evaluate the 3rd integral. Now, for the 3rd integral, we'll move the 3 in front of the integrals a constant. That will give us 3, multiplied by the integral from 4 to 6 of X divided by X2 + 9. We are going to use a U substitution in the denominator, allowing you to equal to X2 + 9, then DU will equal to 2 XDX. We have an XDX in this integral, but we need to divide over the two coefficient, leaving us with 1/2 DU is equal to XDX. Then if we apply this substitution, we get 3 halves. Multiplied by the integral from A to B, because these are integrations in terms of you, of 1 divided by U D U. And this is going to simplify to be 3 halves times the natural logarithm of you, evaluated from A to B, and once we backs substitute X, we will have 3 halves multiplied by the natural logarithm of X2 + 9, and this will be evaluated from 4 to 6. So this is going to be the value of the 3rd integral in the sum. Now, each of these values are being evaluated from 4 to 6, so we can go ahead and recombine these back into the overall sum. That is going to leave us with 4/3 excubed. Plus 3 multiplied by the natural logarithm of X, plus 3 halves multiplied by the natural logarithm of X2 + 9, and everything here is going to be evaluated from 4 to 6. Now, what we are going to do is we are going to plug in the values of the bounds. That is going to leave us with the quantity of 4/3 multiplied by 6 cubed, plus 3 multiplied by the natural logarithm of X, plus 3 halves, oh sorry, we're plugging in 6, so that's not natural logarithm of X, that's natural logarithm of 6. Plus 3 halves multiplied by the natural logarithm of 6 squad plus 9. Then we will subtract the value of the lower boundary, which is going to be 4/3, multiplied by 4 cubed. Plus 3, multiplied by the natural logarithm of 4. Plus 3 halves multiplied by the natural logarithm of +42 + 9. Now, by simplifying and combining like terms, this is going to simplify further to be 608 divided by 3, plus 3 halves multiplied by the natural logarithm of 83 divided by 20, and this is going to be the simplest form of the solution to the definite integral. And with that being said, the overall solution to this problem is going to be D. 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.15. ∫ (from 1 to 2) (3x⁵ + 48x³ + 3x² + 16)/(x³ + 16x) 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.
15. ∫ (from 1 to 2) (3x⁵ + 48x³ + 3x² + 16)/(x³ + 16x) dx