71. Different Methods
Let I = ∫ (x²)/(x + 1) dx.
b...

Question

71. Different Methods

Let I = ∫ (x²)/(x + 1) dx.

b. Evaluate I by first performing long division on the integrand.

Accepted Answer

Welcome back, everyone. In this problem, we want to find the integral of X2 + 2X divided by X minus 2 with respect to X by using long division on the integran. Now, for starters, if we're going to use long division and the integrand, then we're going to end up with the result plus a remainder plus some remainder if it has one that we can use and then integrate term by term. So let's go ahead and do that long division. So that means we're going to try and find out. How many times X minus 2 can go into X2 + 2 X? No, remember for a long division here we're going to put 0 to represent our constant, OK, since there's no constant here, and to do it, we are going to take the leading term of the divisor and put it into each term of the dividend. So for starters, X goes into X2 X times, and then if we multiply X by X minus 2, we're going to end up with X2 minus 2 X. And when we subtract that from our dividend, we'll have X minus X 0 and 2 X minus negative 2X to give us 4 X. Next, X goes into 4 x 4 times, so we put + 4 here in our result, and then we bring down our + 0 because we're going to multiply X minus 2 by 4 and subtract it. So that would be 4 X minus 8 and 4 X minus 8 subtracted from 4 X + 0 would leave us with positive 8. No, in this case x wouldn't go into it, so here we are left with our remainder. It is our remainder. So thus, OK, this tells us then that the integral of X squared plus 2 x divided by X minus 2 with respect to X. Is going to be equal to the integral of x + 4 + 8 divided by x minus 2 with respect to x. So now we can go ahead and integrate term by term. Now let's start with our first term. So no integrating term by term. Let's first integrate X. Now the integral of X with respect to X is going to be equal to a half of X2 plus a constant. Let's call this constant C1. Next, the integral of 4 with respect to X is going to be equal to 4 X plus some constant C2. Now for the integral, let me use a straight line here for the integral of and let me make some space actually coming up, right? For the integral of 8 divided by X minus 2 with respect to X we can integrate this with substitution. Now if we let you. Be equal to X minus 2, we'll be able to substitute you. And if we differentiate you with respect to X, then it will be equal to 1, which implies that DU equals DX. OK? So we can substitute DX for a DU. So now by substitution, then the integral of 8 divided by X minus 2. With respect to X. Is going to be equivalent to the integral of a divided by u with respect to you if we factor out or if we take out our constant here, that's 8 multiplied by the integral of 1 divided by UDu and now we know that the integral here. Would be equal to a multiplied by the natural log of the absolute value of u plus some constant c3, and since we know U equals x minus 2, we can put that in our result so that's 8 multiplied by the natural log of the absolute value of X minus 2 plus C3. Now that we've integrated all three terms, we can combine the results, OK. So now by combining the results. OK. Then this means that we're going to end up with a half X2 plus C1 for our first term plus 4 X + C2 for a second term, plus 8 multiplied by the natural log of the absolute value of X minus 2 + C3 for the third term. And now, since we have 3 constants, we can combine all of them into one big constant. So this is gonna be a half of X squared plus 4 X plus 8 multiplied by the natural log of the absolute value of X minus 2 plus C. This is the integral of X2 + 2 X divided by X minus 2 with respect to X. Thanks a lot for watching everyone. I hope this video helped.

71. Different Methods
Let I = ∫ (x²)/(x + 1) dx.
b. Evaluate I by first performing long division on the integrand.

Key Concepts

Long Division of Polynomials

Integration Techniques

Definite vs. Indefinite Integrals

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