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.

∫ [(x³ + 1) / (x³ − x)] dx

Accepted Answer

Welcome back everyone. In this problem we want to evaluate the integral of x + 8 divided by x minus 4 x with respect to x. Now to evaluate this integral, notice that the degree of the numerator 3 is equal to the degree of the denominator 3. So in this case we have to use polynomial long division to help us simplify before applying partial fraction decomposition. So let's go ahead and divide x cubed plus 8 by x cubed minus 4 x. Now, to do that, OK, first notice that x cubed goes into x cubed 1 time, and if we multiply 1 by X minus 4 x, we get X cubed minus 4 x. And know when we subtract x minus X is 0. And then we'll have negative -4X, which is a positive 4X plus 8. So x + 8 divided by x minus 4 x. Equals 1 + 4 x + 8 divided by x cubed minus 4 x. Now that we have our integral this way, we can set up our partial fraction decomposition for the remainder term. Now notice here, OK. That if we take 4 x plus 8 divided by x cubed minus 4 x, we could rewrite it as 4 x plus 8 divided by x multiplied by x minus 2 multiplied by x plus 2 because here x cubed, let me put this in a box to our right here x cubed minus 4 x could be rewritten as x multiplied by x2 minus 4. And x2 minus 4 is the difference of 2 squares, so we could rewrite that as x multiplied by x minus 2 multiplied by x + 2. So that's how we got it in this form. And when we have it this way, then our partial fractions will be a divided by x plus B divided by x minus 2 plus C divided by x + 2. So let's see if we can figure out the values of A, B, and C for our partial fractions. Now if we multiply both sides by the common denominator, then we get 4 x + 8 to be equal to a multiplied by x minus 2 multiplied by x + 2. Plus b multiplied by x multiplied by x + 2. Plusc multiplied by x multiplied by x minus 2 and now to find the constants, let's substitute the roots of the denominator first to solve for a, OK, we can let x be equal to 0. Then in that case. 4 multiplied by 0 + 8 would be equal to. OK, would be equal to a multiplied by 0 minus 2 multiplied by 0 + 2, which means 8 equals -4A, so a must be equal to -2. Next, for B, let's let X be equal to 2. In that case, 4 multiplied by 2 + 8 would be equal to b multiplied by 2 multiplied by 2 + 2. OK, so this tells us then that. 16, OK, is going to be equal to 8B, so B would be equal to 2. And finally, for C, let's let X equal. 2 Because if X equals -2, then we will have 4 multiplied by -2 + 8. Equal to c multiplied by -2 multiplied by -2 minus 2. OK. Which means then that -8 + 8 or 0 will be equal to 8c, so that means that c must be equal to 0. So this tells us then if we go back to our original expression. That the integral of x cubed plus a divided by x cubed minus 4 x with respect to x. Can be rewritten OK as the integral. Of 1 minus 2 divided by x, OK, plus 2 divided by x minus 2, OK, plus 0 divided by x plus 2, which would just be 0, all with respect to x, and now we can integrate each of these terms with respect to x. The integral of 1 is x. The integral of 2 divided by x would be 2 multiplied by the natural log of the absolute value of x. And the integral of 2 divided by x minus 2 would be 2 multiplied by the natural log of the absolute value of x minus 2 plus c. Using what we know about logs and powers, then we can combine both of these natural log terms here to rewrite this as x + 2 multiplied by the natural log of the absolute value of x minus 2 divided by x plus c. Therefore, this is 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.∫ [(x³ + 1) / (x³ − x)] dx

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Evaluate the integrals in Exercises 9–28. It may be necessary to use a substitution first.
∫ [(x³ + 1) / (x³ − x)] dx