7–64. Integration review Evaluate the following integrals.
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Question

7–64. Integration review Evaluate the following integrals.

62. ∫ (-x⁵ - x⁴ - 2x³ + 4x + 3) / (x² + x + 1) dx

Accepted Answer

Hello. In this video, we are going to be evaluating the following integral. We are given the integral of X 5th, minus 2 x 4th, minus X cubed, minus 4 X2 + 4 X + 5, and this is divided by the quantity X2 + x + 2. Now, the first thing that we can do in order to simplify this function is we can perform polynomial long division. Let's go ahead and set up the polynomial long division. That is going to be the top polynomial, which is X 5th, minus 2 X 4th, minus X cubed, minus 4 X2 + 4 X + 5, and this is going to be divided by the quantity X2 + X + 2. Now, we are going to multiply this polynomial by a term that is going to allow us to cancel the first term X 5th. If we were to multiply that term by X cubed, then this will give us X 5th power, plus x 4th power plus 2 X cubed. Then we will need to subtract this quantity afterward. That is going to allow us to cancel the X of the 5th term, and we are left with negative 3x the 4th minus 3X cubed, and once we bring down the next term, we have -4 X squared. Now, if we were to multiply our polynomial by -3X2d, that is going to leave us with negative 3x the fourth minus 3X cubed minus 6 X2. And again, we will need to subtract this quantity afterward. That is going to allow us to cancel out the X the 4th and X cube term, and we will be left with positive to X squared. Now, because we need to simplify trinomial with a trinomial, we will bring down the next two terms. Now, finally, if we were to multiply our polynomial, our denominator, by let's say, 2, then we will end up with 2 X2 + 2 X + 4. And again we will subtract this quantity. And that will allow us To eliminate the X square terms, leaving us with 2 X plus 1 as a remainder. Now by polynomial long division, this has allowed us to rewrite the original integral to be the integral of our quotient, which is X cubed minus 3 X2 + 2 plus the remainder, which is 2 X + 1 divided by the original denominator, which is X2 + X + 2, and we will integrate with respect to X. Let's go ahead and group up the first two terms together and split this into a sum of integrals. We will call this integral 1 and integral 2, and we will integrate each of them separately. Starting with integral 1, if we were to take the antiderivative each term independently, we will end up with 14th X to the 4th power minus X cubed plus 2 X. This will be the antiderivative of the first integral. But what about the second integral? Now, if we were to allow or apply a substitution, let's say U is equal to our denominator, X2. Plus x + 2, then DU is equal to 2 x + 1 d X, which we do have in the current integral. So this is going to allow us to simplify that integral to be the integral of 1 divided by UDU, which simplifies to the natural logarithm of U. Then we can resubstitute this natural logarithm back into terms of X, and this will be the natural logarithm of X2 + x + 2. And so now, if we were to recombine the terms together, we will end up with 1/4 X 4th power, minus X cubed plus 2 X plus the natural logarithm of X2 + X + 2. And because we were working with a general antiderivative, we will add on a general constant C. And this is going to be the solution to the original integral. And with that being said, the solution to this problem is going to be B. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

7–64. Integration review Evaluate the following integrals.
62. ∫ (-x⁵ - x⁴ - 2x³ + 4x + 3) / (x² + x + 1) dx

Key Concepts

Integration

Polynomial Long Division

Rational Functions

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