In Exercises 33–38, perform long division on the integrand, wr...

Question

In Exercises 33–38, perform long division on the integrand, write the proper fraction as a sum of partial fractions, and then evaluate the integral.

∫ 2y⁴ / (y³ - y² + y - 1) dy

Accepted Answer

Hello. In this video we are going to be evaluating the following integral. We are given the integral of 3 x 4 divided by x cubed plus x2 + x + 1. Now, since the denominator's power is less than the numerator's power by exactly one, we can go ahead and simplify the integrand by using polynomial long division. So let's go ahead and set up the polynomial long division. That is going to be the denominator x cubed plus x2 + x + 1, and we are going to divide the numerator, which is 3x the 4th power. Now, what we want to try to do is match this first term to the current term by multiplying by a coefficient or variable. Now, in order to change X cub to 3 x 4th, we need to multiply it by 3 X. So we're going to multiply the entirety of the denominator by 3 X and then add subtract that from the from the denominator, or subtract that from the quotient. So that is going to give us 3 x 4th plus 3 X cubed + 3 X2 + 3 X, and keep in mind that we have to subtract this quantity afterward, so distributing the negative sign is going to cause all these values to be negative. So now we will go ahead and combine like terms. The 3 x to the 4th terms will cancel each other out, and that will leave us with -3 X cubed minus 3 X 2 minus 3X. Now we can still match the first term of our denominator X cubed to -3 X 2 by multiplying by 3. Multiplying the denominator by -3 will give us -3 x cubed minus 3 x 2. -3 x minus 3. And then once we distribute our negative sign for the subtraction, all these terms are going to change to positive terms. So by combining like terms, 3 X cube terms cancel each other out, 3 X 2 terms cancel each other out, and the 3 X terms cancel each other out, leaving us for the remainder of 3. So what this means is that now we can rewrite. The integral by using the polynomial long division, and that will allow us to rewrite the integral to be the integral of 3 x minus 3 + 3 divided by x cubed plus X2 + X + 1. And since we have a sum or difference of three terms, we can assign each of these their own integral symbol. Now, what we want to do is we want to go ahead and evaluate each of these integrals independently. We will start with the first two integrals. Now, the first integral is fairly easy to solve, because the integral of 3 X is going to simplify to be 3 halves X squad. And for the second integral, we have the negative integral of 3, but because this is a constant, the value of this integral is going to be -3 X. So this is going to be the results of the first two integrals, but the last integral is going to be a little trickier to solve. So, when approaching the 3rd integral, the first thing we want to do is simplify the numerator. Now we can simplify the numerator by using factory by grouping. Grouping the first two and last two terms together, notice that we can factor out an x 2 from the first group that will allow us to rewrite the denominator to be x2 multiplied by x + 1 plus the quantity x + 1. And combining like terms together allows us to rewrite the denominator as X2 + 1 multiplied by X + 1. So what this means is that we can rewrite the integrant to be 3 divided by X2 + 1 multiplied by X + 1, and we can further simplify this by using partial fraction decomposition. Now here we have a product of a square term and a linear term. So our respective partial fractions is going to be a divided by x + 1 plus B X + C divided by X 2. Plus one And so now what we want to do is we want to solve for the coefficients of the partial fractions. We will multiply everything by the lowest common denominator, and that is going to give us 3 is equal to a multiplied by x2 + 1 plus the quantity b x + c multiplied by x + 1. Now, if we were to multiply and reorder the terms, we can rewrite this equation to be the quantity 3 equal to A + B multiplied by X2, plus B + C multiplied by X plus the constant terms A + C. Now we can create a system of linear equations by setting the coefficients equal to each other. Setting the coefficients of the x 2 terms equal to each other gives us a plus B equal to 0. For the second equation, setting the coefficients of the x terms equal to each other gives us B + C equal to 3, and setting the coefficients or the constant terms equal to each other gives us the third equation, A plus C equal to 3. Now, if we were to solve this system, we will get the following coefficients. We get A's equal to 3 halves. B is equal to -3 halves. And c is equal to 3 halves. And so now what we want to do is we want to take these coefficients and plug them back into the partial fractions. So, that is going to give us the partial fractions, 3 halves, divided by X plus 1. Plus -3 halves. Multiplied by x plus 3 halves divided by x2 + 1. So this is going to be the way that we rewrite the 2nd integral. Now let's go ahead and clear some room here. OK, so, we've rewritten the integrand of the 3rd integral by using partial fraction decomposition, but now what we can go ahead and do is further split up the second part of the sum by making it a sum of fractions. So we can further simplify this result to be 3 halves divided by X plus 1. -3 halves divided by x2 + 1. With an X in the numerator, plus 3 halves divided by X2 + 1. And now what we can go ahead and do is we can assign each of these fractions their own integral, and now we have to simplify each of these independently. So, starting with the first integral, we have 3 halves, X plus 1. Moving that coefficient to the front gives us 3 halves multiplied by the integral, 1 divided by X plus 1, and this is going to simplify to be 3 halves multiplied by the natural logarithm of X plus 1. So this is going to be the result of the first fraction. Now, if we take a look at the 2nd integral, moving the 3/2 coefficient to the front gives us -3 halves multiplied by X divided by X2 + 1. We can simplify this by using a U substitution and allowing you to equal to x2 + 1. DU will equal to 2 XDX, which can also be rewritten as 1/2 DU equal to XDX. Then this integral will simplify to be -3/4 multiplied by the integral of 1 divided by UDU, which will further simplify to be 3/4 multiplied by the natural logarithm of u. And by resubstituting or substitution of u, we get 3/4 multiplied by the natural logarithm of x2 + 1. This is going to be the result of the 2nd integral. And for the final integral. We will have Three halves Multiplied by the integral of 1 divided by x2 + 1 and by using our tangent inverse property this will integrate to be 3 halves multiplied by the tangent inverse of x which is going to be the result of the 3rd integral. So these are going to be the three results of the independent integrals, and keep in mind that we also have 3 halves X2 and -3 X as part of our solution. So by combining all of the results together, we will get a very lengthy solution that is going to be 3 halves X2 minus 3X plus 3 halves multiplied by the natural logarithm of X + 1. 3/4 multiplied by the natural logarithm of x2 + 1. Plus 3 halves multiplied by the tangent inverse of x and because we were originally working with an indetermined. An undefinite integral. We will add a generic constant C. This is going to be the solution to the entire integral. And with that being said, the overall solution to this problem is going to be option C. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

In Exercises 33–38, perform long division on the integrand, write the proper fraction as a sum of partial fractions, and then evaluate the integral.∫ 2y⁴ / (y³ - y² + y - 1) dy

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In Exercises 33–38, perform long division on the integrand, write the proper fraction as a sum of partial fractions, and then evaluate the integral.
∫ 2y⁴ / (y³ - y² + y - 1) dy