Evaluate the integrals in Exercises 9–28. It may be necessary ...

Question

Evaluate the integrals in Exercises 9–28. It may be necessary to use a substitution first.

∫ [(2x³ + x² − 21x + 24) / (x² + 2x − 8)] dx

Accepted Answer

Welcome back everyone. In this problem we want to evaluate the integral of XQ + 3 X2 minus 4 X + 5, all divided by X2 + X minus 2 with respect to X. Now, to evaluate this integral, we must first address the fact that the degree of the numerator is higher than that of the denominator. In other words, notice here that the degree of our numerator is 3, while the degree of our denominator is 2. So we'll need to use polynomial long division to simplify our quotient and then further simplify our remainder using partial fraction decomposition. That's when we'll be able to integrate. So let's go ahead and do that, OK? So let's start by dividing x cubed plus 3X2 minus 4 X + 5. By x2 + x minus 2 No, X2 goes into x cubed x times, OK, and if we multiply x by x2 + x minus 2, we end up with x cubed plus X2 minus 2x. OK, so now we can subtract this from. XQ + 3 X2 minus 4 x + 5, and when we do that, we end up with 2 X2 minus 2 x + 5. Now x2 goes into 2 x 22 times, and when we multiply through by 2, we end up with 2 x 2 plus 2 x minus 4. And now if we subtract that, OK, then we're left with -4 x plus 9. So this tells us that our integran. Can be rewritten as x + 2 plus or remainder of -4 x + 9 divided by x2 + x minus 2. Now we can set up the partial fraction decomposition for our remainder. Now, focusing on our remainder here, so let me make a note of that. or rather let me say decomposing. Our remainder. Into partial fractions, OK. Then what we're saying is that we want to rewrite -4 x + 9 divided by X2 + X minus 2, OK. Which of course we can rewrite our denominator as the product of two linear terms, the product of x + 2 and x minus 1. And now we want to rewrite this as A divided by X + 2 + B divided by X minus 1, where we're trying to find the values of A and B. Now if we multiply both sides by X + 2 multiplied by X minus 1, then we should get -4 x + 9 to be equal to A multiplied by X minus 1. Plus b multiplied by x + 2. No, to find B, let's let x be equal to 1. Then -4 multiplied by 1 + 9 will be equal to b multiplied by 1 + 2. And if that's the case, then that means that 5 will be equal to 3b. So b will be equal to 5/3. Next, to find a, we can let X, OK, be equal to -2. And now in this case that means 4 multiplied by -2 + 9 will be equal to a multiplied by -2 minus 1. This means 17 equals -3A, so A would be equal to -173. So if we substitute the quotient and the partial fractions back into the integral, OK, then that means our integral of Well, let me write it here or integral of x cubed. Plus 3 x2 minus 4 x + 5 divided by x2 + x minus 2 with respect to x OK will be equal to the integral. Of x + 2 minus 17 divided by 3 multiplied by x + 2 + 5 divided by 3 multiplied by x minus 1, all with respect to x. And now we can integrate each of these terms separately. When we do that, the integral of x will be 12 of x 2. The integral of 2 is 2x. The integral of 17 divided by 3 multiplied by x + 2 will be -17/3 of the natural log of the absolute value of x + 2. While the integral of 5 divided by 3 multiplied by x minus 1 will be 5/3 of the natural log of the absolute value of x minus 1 plus our constant c, thus this would be our integral. Thanks a lot for watching everyone. I hope this video helped.

Evaluate the integrals in Exercises 9–28. It may be necessary to use a substitution first.∫ [(2x³ + x² − 21x + 24) / (x² + 2x − 8)] dx

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Evaluate the integrals in Exercises 9–28. It may be necessary to use a substitution first.
∫ [(2x³ + x² − 21x + 24) / (x² + 2x − 8)] dx