2–74. Integration techniques Use the methods introduced in Sec...

Question

2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.

71. ∫ (2x² - 4x)/(x² - 4) dx

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to evaluate the following integral, and we're given the integral of the quantity of 3 X2 minus 6 X in quantity, divided by the quantity of X2 minus 9 in quantity DX. So the first step in evaluating this integral is to simplify our integram. So what we're going to do is take our denominator, which is X2 minus 9, and divide it into our numerator of 3 X2 minus 6 X. We're going to do this by long division. So, X2 will go into 3 X2 3 times. And so then we'll have 3 multiplied by the quantity of X2 minus 9, and that will give us 3 X2 minus 27. And then we're going to want to subtract that quantity from 3 X quad minus 6 X. In doing so, we'll see that we have a remainder of minus 6 X plus 27. So that means that are integral. Of the quantity of 3 X2 minus 6 X in quantity, divided by the quantity of X2 minus 9 in quantity DX is going to be equal to the integral of 3DX. Plus the integral of the quantity of minus 6 x plus 27 in quantity divided by ad here for a denominator instead of writing this as X2 minus 9, we're going to actually factor that since it's the difference of two squares. So we'll have the quantity of X plus 3 in quantity multiplied by the quantity of X minus 3 in quantity DX. So now what we want to do is to take that 2nd interval and break it up into two separate fractions using partial fractions. So that means that we will have minus 6 X plus 27, and that quantity is going to be divided by the quantity of X + 3 in quantity, multiplied by the quantity of X minus 3 in quantity, and then that fraction is going to be equal to, and here we'll use partial fractions. So we'll have a divided by the quantity of X plus 3. Plus B. Divided by the quantity of x minus 3 in quantity. So multiplying both sides of this equation by the quantity of X + 3 in quantity multiplied by the quantity of X minus 3 in quantity, we will get that we have minus 6 X plus 27 is equal to a multiplied by the quantity of X minus 3 in quantity, plus B multiplied by the quantity of X plus 3 in quantity. And so now we just need to equate the coefficients in both the X variable as well as the constant terms. So looking at the X variable, we will have that minus 6 is equal to a plus B, and if we look at the constant terms, we'll have that 27 is equal to minus 3A plus 3B. And for the second equation, we can divide both sides of the equation by 3. In doing so, we'll get that 9 is equal to minus A plus B. So now we can take our two equations here and add them together. In doing so, we will get that 3 is equal to 2 multiplied by B, and dividing both sides of the equation by 2, we'll get that B is equal to 3 divided by 2. And we can substitute this result back into either equation. And so we will substitute back into the first equation, which is gonna be minus 6 is equal to a plus 3 divided by 2, and we can solve this for A by subtracting 3 divided by 2 from both sides. In doing so, we see that A is going to be equal to minus 15 divided by 2. So this means that our original interval, which is going to be the integral. Of the quantity of 3 x 2 minus 6 X in quantity, divided by the quantity of X2 plus 9 in quantity DX is going to be equal to, we still have the first term, which is going to be the integral of 3DX. Then we'll have two additional integrals. So we'll have plus. The integral, and here we'll have A, which is the quantity of minus 15 divided by 2 in quantity, and that gets divided by the quantity of X plus 3 in quantity DX. And then for the next integral, we'll have plus the integral. Of the quantity, and here we'll have B, which is 3 divided by 2, in quantity, divided by the quantity of X minus 3, in quantity DX. So now we just need to perform the integration. So performing each integral individually, this is going to be equal to. For the first integral, we'll have just the interval of 3 with respect to X, and that's just going to give us 3 X. Then we'll have for the next interval, we'll go ahead and pull out the constant in front of the interval. So we'll have minus 15 divided by 2, and this gets multiplied by the interval of 1, divided by the quantity of X + 3 in quantity DX. and recall that that's just going to be equal to the natural log. Of the absolute value of X plus 3 in quantity. Then we'll have plus for the next interval. We'll go ahead and pull out the constant in front of the integral, so that'll be 3 divided by 2, and that gets multiplied by the integral of 1 divided by the quantity of X minus 3 and quantity DX. And so performing that integration, we will get the natural log of the absolute value of x minus 3 in quantity, and then we'll have to add our integration constant of plus C. So now we just need to simplify this expression, so we can factor out a 3 divided by 2 out of all of the terms. In doing so, this is going to be equal to 3 divided by 2, multiplied by the quantity of 2 X. Minus 5 multiplied by the natural log of the absolute value of the quantity of X + 3 in quantity, plus the natural log of the absolute value of the quantity of X minus 3 in quantity, then another in quantity and plus C. So now we want to do is combine our two natural log functions using the properties of logarithms. In doing so, we see that our integral is going to be equal to 3 divided by 2, multiplied by the quantity of 2 X. Plus the natural log of the absolute value of the quantity of the quantity of X minus 3 in quantity divided by the quantity of X plus 3 in quantity raised to the 5th power in quantity, then another in quantity and plus C. So 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.

2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.
71. ∫ (2x² - 4x)/(x² - 4) dx

Key Concepts

Partial Fraction Decomposition

Polynomial Division

Basic Integration Techniques

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