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.

∫ 1 / (x⁴ + x) dx

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 1 divided by x 4 + 9 x dx. OK, so it appears for this particular prompt, we're asked to evaluate the integral that is provided to us. So now that we know that we're ultimately trying to evaluate this integral based on our prior knowledge, in order to evaluate the integral of 1 divided by x 4 plus 9 x dx, just by looking at the integral. Itself, we should be able to recall that we can use algebraic manipulation followed by a substitution and a partial fraction decomposition in order to evaluate this integral. So we're going to use a little bit of algebra at first to simplify our integral to make it easier to manipulate to evaluate at the very end. So, We have to just break it up into parts so we can make this problem as easy to solve as we can. So, with that said, our first step now is we need to use our algebraic manipulation to help us factor out the denominator. So in order to do that, we must first factor out an X from the denominator, so we could take x to the power of 4. Plus 9 x is equal to x multiplied by parentheses of x to the 3 plus 9, so that will mean our integral will become i is equal to the integral of 1 divided by x multiplied by parentheses of x to the 3 + 9 dx. Fantastic. Our 2nd step now is we now need to use some algebraic manipulation and apply our substitution method. And in order to make our substitution method a little bit easier to apply, we will need to multiply the numerator and denominator by x 2. So when we do that, we can simply write i is equal to the integral of x 2. Divided by x 3 multiplied by x 3 plus 9 in parentheses. D X Now, let's make a note. Now we can let you be equal to x 3. Then we could write that DU by DX, so DU by DX will be equal to 3x to the power of 2. Which will mean that X 2DX will be equal to 1/3 DU. Fantastic. So, moving right along here, so that will mean we can substitute these values into our integral, so we could write that i is equal to 1/3 multiplied by the integral of 1 divided by u multiplied by parentheses of u plus 9 DU. Fantastic. So now our 3rd step is we need to recall and apply our partial fraction decomposition method. So we will need to decompose the integrand, 1 divided by U multiplied by parentheses of U plus 9. And this will be equal to A divided by U plus B divided by U plus 9. So, when we multiply by the common denominator, which in this case the common denominator is U multiplied by parentheses of U + 9, We will be able to write that 1 is equal to a multiplied by parentheses of U + 9 + B multiplied by U. OK. So, Now we need to set u is equal to 0, and when U is equal to 0, we could write 1 is equal to 9 multiplied by A, and when we rearrange this expression to isolate and solve for A, we will find that A is equal to 1/9, and once again you're just using algebra to move the equation around just so you could solve for A by itself. And when U is equal to -9, similarly, we could write that 1 is equal to -9 multiplied by B, which will mean that B is simply equal to 19, whereas A is equal to 19. Which will mean our integral will become. I is equal to 1/3 multiplied by the integral of parentheses of 1/9 divided by U minus 1/9 divided by you plus 9. Do you? Which will be equal to 1/27, 1 divided by 27 multiplied by the interval of parentheses of 1 divided by U minus 1 divided by U plus 9DU. Fantastic. So now for our fourth and final step is we need to perform our integration and back substitution. So we need to integrate the term, so i is equal to 1/27 multiplied by parentheses of the natural log of the absolute value of u minus the natural log of the absolute value of u plus 9. Plusc where c denotes some constant variable. And when we recall and use the logarithmthic quotient rule, you can go ahead and write that i is equal to 1/27 multiplied by the natural log of the absolute value of u divided by u plus 9. Plus C. And finally, when we back substitute that u is equal to x 3, we could write that our final answer will be i is equal to 1/27 multiplied by the natural log of the absolute value of x 3 divided by x 3 plus 9 and absolute value signs plus c. And that's it. That is our final answer. Hooray, we did it. So with that in mind we could safely say once again that our final answer for our evaluated integral will be once again 1/27 or 1 divided by 27 multiplied by the natural log of the absolute value of x to the power of 3 divided by x 3 plus 9 plus 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.∫ 1 / (x⁴ + x) dx

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In Exercises 21–32, express the integrand as a sum of partial fractions and evaluate the integrals.
∫ 1 / (x⁴ + x) dx