23-64. Integration Evaluate the following integrals.
32....

Question

23-64. Integration Evaluate the following integrals.

32. ∫ (4x - 2)/(x³ - x) dx

Accepted Answer

Welcome back everyone. In this problem, we want to find the integral of 6 X minus 4 divided by X cubed minus X with respect to X. Now let's try to solve this integral using partial fractions. First, let's factor or denominator. Now if we think about it, X cubed minus X is equal to x multiplied by x2 minus 1, which could be rewritten as x multiplied by x minus 1 multiplied by x + 1 based on the difference of 2 squares. So in that case, we could say then. That 6 x minus 4 divided by X cubed minus X is going to be equal to 6 X minus 4 divided by X multiplied by X minus 1 multiplied by X + 1. Which we can break up as a partial fraction as a divided by X plus b divided by X minus 1 plus C divided by X plus 1. Now that we have that, let's try to solve for A, B, and C. Let's multiply both sides by X multiplied by X minus 1, multiplied by X + 1. When we do that, then we get 6 X minus 4 to be equal to A multiplied by X minus 1, multiplied by X + 1 plus B X B multiplied by X. Multiplied by X + 1. Plusc multiplied by X multiplied by X minus 1. Now if we expand some of our terms here. We are going to have a multiplied by X2 minus 1 plus B multiplied by X2 + X. Plus C multiplied by x2 minus X. I know if we were to rearrange here in terms of our X terms, OK. For X2d terms, we're going to have A + B + C. So all of that multiplied by X2. Plus, for our X terms, B minus C. OK, so that's B minus C multiplied by X minus A, because A multiplied by -1 would give us negative A. So all of this is equal to 6 X minus 4. Now if that's the case, we can match the coefficients on the left and right-hand sides of our equation. In other words, for the X coefficient, there's no X coefficient on the left, so that's 0, and on the right, that it's, that's going to be A + B + C. So A + B + C equals 0. For our X term 6 coefficient of X on the left will be equal to b minus C. It's coefficient on the right, and for its constant. Then our constant -4 will be equal to -8. Now there are a few things that we can tell here. First, from our constant, this means that a must be equal to 4. And if A equals 4. OK. Then that means if we substitute it into our formula for X or Xqua coefficients, then 0 will be equal to 4 + B + C. Or B + C will be equal to -4. If B + C equals -4, we can substitute that into our equation, OK, to solve for our system. Because if we add equations, OK, if we're adding equations. OK. Then this tells us that If B minus C equals 6 and B + C equals 4, then B + C plus B minus C must be equal to -4 + 6. In that case, C minus C equals 0, B plus B equals 2B, and if 2 B equals -4 + 6 or 2, then B must be equal to 1. And if B equals 1, this means then. That C +1, if C + B equals -4, C + 1 will be equal to -4, which means that C equals -5. So with all of this information, OK, now that we've figured out a. we figured out that A is 4, B is 1, and C is -5. Then decomposing our expression, this must mean that 6 X minus 4 divided by X cubed minus X is equal to 4 divided by X. Plus 1 divided by X minus 1. OK. minus 5 divided by X + 1. So for our inte integral then or integral. Of 6 X minus 4 divided by X cubed minus X with respect to X is going to be equal to the integral of 4 divided by X with respect to X plus 1, the integral of 1 divided by X minus 1 with respect to X minus the integral of 5 divided by X + 1 with respect to X. Now we can go ahead and integrate each term. Now the integral of 4 divided by X with respect to X will be equal to 4 multiplied by LN of absolute value of X. The integral of the integral of 1 divided by X minus 1 with respect to X would be equal to LN of the absolute value of X minus 1, and for a third term it would be minus 5 multiplied by LN of the absolute value of X + 1 + C. Thus, this is the integral of 6 X minus 4 divided by X cubed minus 3X minus X sorry, with respect to X. Thanks a lot for watching everyone. I hope this video helped.

23-64. Integration Evaluate the following integrals.32. ∫ (4x - 2)/(x³ - x) dx

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23-64. Integration Evaluate the following integrals.
32. ∫ (4x - 2)/(x³ - x) dx