In Exercises 21–32, express the integrand as a sum of partial ...

Question

In Exercises 21–32, express the integrand as a sum of partial fractions and evaluate the integrals.

∫ (s⁴ + 81) / (s(s² + 9)²) ds

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Evaluate the integral, the integral of x 4 plus 2 x 2 + 16, all divided by x multiplied by the parentheses of x 2 + 4 to the power of 2 d x. OK. So it appears for this particular prong we're simply asked to evaluate the integral that is provided to us. And based on our prior knowledge in order to evaluate the integral. I is equal to the integral of x to the power of 4. Plus 2 x to the power of 2 plus 16 all divided by x multiplied by parentheses of x to the 2 plus 4 to the power of 2 d x. So once again, specifically this interval that's given to us by the prong itself, we need to be able to recall based on our prior knowledge and recognize that looking at this specific interval. We can recall and use the partial fraction decomposition for a rational function with a repeated quadratic factor. So keeping that in mind, our first step that we can take is we could begin to set up our decomposition. So we need to be able to recognize that the denominator has a linear factor X and a repeated irreducible quadratic factor, and let's make a note of this in blue parentheses x to the power of 2. Plus 4, all to the power of 2, and the partial fraction form can be written as. X 4 + 2 x 2 + 16 all divided by x multiplied by parentheses of x 2 + 4 to 2, which will be equal to A divided by x plus B multiplied by x plus C. Divided by x 2 + 4. Plus D multiplied by X plus E. Divided by parentheses of, so to be divided by parentheses of x to the power of 2. Plus 4. All to the power of 2. Fantastic. So now for our second step, so this is step 2, we now need to solve for the constants. So in order to solve for the constants, we must first multiply both sides by the common denominator, which the common denominator is x multiplied by x to the power, so specifically x multiplied by parentheses of x 2 + 42. So we're gonna be multiplying both sides of the common denominator by this expression. And when we do that, we will get x 4 + 2 x 2 + 16, which will be equal to a multiplied by parentheses of x 2 + 4 in parentheses to the power of 2. Plus parentheses of b multiplied by x + c multiplied by x multiplied by parentheses of x to the power of 2 + 4 + parentheses of d multiplied by x. Plus E multiplied by X. So now, in order to find A, so now we're solving for the variable A, we need to let X be equal to 0, and when we do that, we will find that 16 will be equal to a multiplied by 4 to 2. Which will be equal to 16, which is equal to 16 multiplied by a, and when we rearrange the expression to isolate and solve for a by itself using a little bit of algebra, we will find that a is simply equal to 1. So now at this point we need to expand and compare coefficients. So we need to take x 4 + 2 multiplied by x 2 + 16, which will be equal to 1 multiplied by parentheses of x 4 plus 8 multiplied by x 2 + 16. Plus parentheses of B multiplied by x 2 plus c multiplied by x multiplied by x 2 + 4. Plusd multiplied by x to the power of 2. Plus E multiplied by X. Which we can go ahead and go even further to write that x to the power of 4. Plus 2 multiplied by x 2. Plus 16 will be equal to x 4 + 8 multiplied by x 2 + 16 + b multiplied by x 4. Plus. 4 multiplied by b multiplied by x to the power of 2. Plus C multiplied by x to the power of 3. Plus 4 multiplied by C multiplied by X plus D multiplied by X to the power of 2. Plus E multiplied by X. OK, we are on a roll here. So now, at this stage, we need to group the terms by powers of X, so we need to go ahead and write that x 4 + 2 multiplied by x 2 + 16. Will be equal to parentheses of 1 + b multiplied by x to the power of 4. Plus C multiplied by x to the power of 3. Plus 8 + 4 multiplied by B + D multiplied by x 2. Plus 4 multiplied by C + E multiplied by x + 16. So now at this point, we can safely compare the coefficients from both sides. So focusing our attention on the x to the power of 4 coefficient, we could write that 1 minus, so once again 1 will be equal to 1 + B, and when we rearrange this expression to isolate sulfur B, we will find that B is equal to 0. Focusing on her x to the power of 3 coefficient, C will be equal to 0. Looking at our x to the power of 2 coefficient, we will be able to write 2 is equal to 8 + 4 multiplied by 0. Plus D because we know that C is equal to 0, which when we rearrange this expression to isolate and solve for D, you will find that D is simply equal to -6. And finally, looking at our X coefficient, we will be able to write that 0 is equal to or multiplied by 0. Plus E. And when we rearrange this expression to isolate and solve for our E variable, we will find that E is simply equal to 0. So now, with that said. Our simplified integrand will be. X 4 + 2 multiplied by x 2 + 16 all divided by x multiplied by parentheses of x to the 2 plus 4 to 2 will be equal to 1 divided by x minus 6 x divided by parentheses of x to the 2 plus 4 all to the power of 2. Fantastic. We are on a roll here. So, our 3rd step now, this is gonna be step number 3, is we need to integrate our terms, so now we need to integrate each part separately. So when you go ahead and write that i is equal to the integral of 1 divided by x, DX minus the integral of 6 x divided by parentheses of x 2 + 4 to the power of 2 dx. And focusing on our focusing our attention on the first integral, we will be able to take that the integral of 1 divided by XDX will be equal to the natural log of the absolute value of x plus C subscript 1, where C represents some constant variable, and to solve for our second integral, we will need to recall and use the u substitution method. And we need to use the U substitution method on the integral of 6 X divided by parentheses of X2 + 42. DX and using the u substitution on this specific integral, we could say that u is equal to x to the 2 + 4, which will mean that duu by dx will be equal to 2x, which will mean that 3 multiplied by duu will be equal to 6 x dx. So that will mean that the integral of 3 divided by u to the 2DU will be equal to 3 multiplied by the integral of u to the power of -2. D u, which will be equal to 3 multiplied by parentheses of u to the power of 1 divided by -1 plus c subscript 2. Which when we write our final answer in its most simplified form, we will find that it's equal to -3 divided by drum roll x 2 + 4 + C subscript 2. So now for our 4th and final step, we need to combine the results and substitute in back our value of U is equal to x 2 + 4. Fantastic. So when we do that, we'll be able to write that I is equal to the natural log of the absolute. Value of x minus -3 divided by x 2 + 4. Plus C, which will be equal to. I, which is equal to the natural log of the absolute value of x. Plus 3 divided by X2 + 4 + C. So we can combine C subscript 1 and C subscript 2 to just equal C. And that's it. That is our final answer. Hooray, we did it. We saw for our final answer. So going back to look at the prom itself, we could safely say without a doubt that our final answer for this evaluated integral is once again, the natural log of the absolute value of x + 3 divided by x 2 + 4 + C, and that's it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

In Exercises 21–32, express the integrand as a sum of partial fractions and evaluate the integrals.∫ (s⁴ + 81) / (s(s² + 9)²) ds

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In Exercises 21–32, express the integrand as a sum of partial fractions and evaluate the integrals.
∫ (s⁴ + 81) / (s(s² + 9)²) ds