23-64. Integration Evaluate the following integrals.
29....

Question

23-64. Integration Evaluate the following integrals.

29. ∫₋₁² [(5x) / (x² - x - 6)] dx

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to evaluate the integral, and we're given the integral from -1 to 0 of 4 X divided by the quantity of X2 minus 4 X + 3 in quantity DX. So for this problem, we're asked to evaluate the interval, and we're given the interval from -1 to 0 of 4 X divided by the quantity of X2 minus 4 X plus 3 in quantity DX. And we noticed that we have a quadratic in the denominator, and this quadratic can actually be factored. That means that this is going to be equal to the integral from -1 to 0 of 4 X divided by the quantity, and here we'll have the quantity of X minus 1 in quantity multiplied by the quantity of X minus 3 in quantity DX. So now what we want to do is to break up this fraction into two separate fractions using partial fractions. We recall for partial fractions that we will have 4 x divided by the quantity of the quantity of X minus 1 in quantity multiplied by the quantity of X minus 3 in quantity, and this is going to be equal to a divided by the quantity of X minus 1 in quantity, plus B divided by the quantity of X minus 3. So, we need to solve for A and B. The first step is to multiply both sides of the equation by the denominator on the left hand side of our equation. In doing so, we will get that 4 X. is equal to a multiplied by the quantity of X minus 3 in quantity, plus B multiplied by the quantity of X minus 1 in quantity. So multiplying everything out on the right-hand side of this equation, we will have that 4 X is equal to A X minus 3A. Plus BX minus B. and collecting light terms, we will have that 4 X is equal to the quantity of A plus B in quantity, multiplied by X. 3 A minus B. So, equating the like powers of X on both sides of the equation for the linear term, we see that A plus B is going to be equal to 4, and for the constant term, we will see that minus 3A minus B is going to be equal to 0, since we have no constant term on the left hand side. So here we have two equations with two nodes, and so we need to solve them simultaneously. Now, if we add these two equations together, we'll eliminate our variable B. So, in doing so, we will have minus 2A is equal to 4, and dividing both sides of the equation by minus 2, we will get that A is equal to -2. Now, we can substitute this result into either of the equations that we have, so substituting it into the first equation, we will have that minus 2 plus B. is equal to 4, and solving this for B, we'll see that B is going to be equal to 6. So, our original integral, which was the integral from -1 to 0 of 4 X divided by the quantity of X2 minus 4 X plus 3 in quantity DX is going to be equal to the interval from -1 to 0 of -2 divided by the quantity of X minus 1 in quantity DX. Plus the interval from -1 to 0 of 6 divided by the quantity of X minus 3 in quantity DX. So now we just have to integrate both of these intervals. To perform the integration, this is going to be equal to. For the first interval, we'll have -2 multiplied by, and here we're going to be integrating 1 divided by the quantity of X minus 1 in quantity with respect to X, and recall that that will give us the natural log of the absolute value of the quantity of X minus 1 in quantity, and this needs to be evaluated from -1 to 0. Then we'll have plus 6 multiplied by, and for the second integral here, we're going to be integrating 1 divided by the quantity of X minus 3 in quantity with respect to X, and that will give us the natural log of the absolute value of the quantity of X minus 3 in quantity, this needs to be evaluated from -1 to 0 as well. So substituting in or limits of integration, this is going to be equal to. -2, multiplied by the natural log of the absolute value of the quantity of 0 minus 1 in quantity. Plus, 2, multiplied by the natural log of the absolute value of the quantity of -1 -1. In quantity. Then we'll have + 6 multiplied by the natural log of the absolute value of the quantity of 0 minus 3 in quantity. -6 multiplied by the natural log of the absolute value of the quantity of -1 minus 3 in quantity. So now we just need to simplify this expression. So doing so, this is going to be equal to. -2, multiplied by the natural log of 1. Plus 2 multiplied by the natural log of 2. Plus 6, multiplied by the natural log of 3. -6 multiplied by the natural log of 4. Now, the natural log of 1 and 0, so that means that this is going to be equal to 0, plus 2 multiplied by the natural log of 2. Plus, and here we can combine the last two natural log terms using the properties of logarithms. So we'll have 6 multiplied by the natural log of the quantity of 3 divided by 4 in quantity. As combining these two natural log functions, we see that this is going to be equal to. 2, multiplied by the natural log. Of the quantity of 2 multiplied by 3 cubed. And that is going to be divided by. Or You In quantity. And finally simplifying this expression, this is going to be equal to 2 multiplied by the natural log of the quantity of 27 divided by 32 in quantity. And this here is the answer that we're looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

23-64. Integration Evaluate the following integrals.29. ∫₋₁² [(5x) / (x² - x - 6)] dx

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23-64. Integration Evaluate the following integrals.
29. ∫₋₁² [(5x) / (x² - x - 6)] dx