23-64. Integration Evaluate the following integrals.
50....

Question

23-64. Integration Evaluate the following integrals.

50. ∫ 8(x² + 4)/[x(x² + 8)] dx

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says determine the following integral, and we're given the integral of 6 multiplied by the quantity of X2 + 9 in quantity, divided by the quantity of X multiplied by the quantity of X2 + 6 in quantity DX. So the first thing we want to do is to use partial fractions to break up our fraction, that is our integrand, into multiple fractions. And so we call for partial fractions that we will have 6 multiplied by the quantity of X2 + 9 in quantity, and that is going to be divided by the quantity of X multiplied by the quantity of X2 + 6 in quantity, and that this is going to be equal to a divided by X plus the quantity of B X plus C in quantity divided by the quantity of X2 + 6 in quantity. So now we need to multiply both sides of this equation by the denominator on the left hand side. In doing so, we will get that 6 multiplied by the quantity of X2 + 9. In quantity is going to be equal to a multiplied by the quantity of X2 + 6 in quantity. Plus the quantity of bx plus C in quantity multiplied by x. Now multiplying everything out on both sides of the equation, we will get that 6 x squared plus 54 is equal to a x2 plus 6 A. Plus B X 2 plus C X. And collecting like terms, we will have that 6 X 2d plus 54 is equal to the quantity of A plus B in quantity multiplied by X2 plus CX plus 6 A. So matching up the coefficients for like powers of X on both sides of this equation for the quadratic term, we see that A plus B has to be equal to 6. For the linear term, we have that C is equal to 0, since there's no linear term on the left hand side of the equation, and for the constant term, we will have that 6A is equal to 54. Now we can use this last equation here to solve for A, so dividing both sides by 6, we get that A is equal to 9. We can substitute in this value into the first equation to solve for B, so we'll have 9 plus B is equal to 6. So that means that B is going to be equal to. -3. So now we have values for A, B, and C. And so, we can rewrite our original integral, which is the integral of the quantity of 6 multiplied by the quantity of X squared plus 9 in quantity. Divided by the quantity of X, multiplied by the quantity of X squared plus 6 in quantity d X, that this is going to be equal to the integral of 9 divided by X D X plus the integral of minus 3 x divided by the quantity of X2 + 6 in quantity. DX. So now we want to do is to evaluate both of these intervals individually. So looking at the first integral, we have the integral. Of 9 divided by X D X, and we can just integrate this directly. So this is going to be equal to 9, multiplied by, and here we're going to be integrating 1 divided by X with respect to X. And so recall that that's going to give us the natural log of the absolute value of X plus C. Now for the second interval, we will have the interval of minus 3 x divided by the quantity of X2 + 6 in quantity dx. And to evaluate this interval, we're going to need to make a substitution. So we're going to set U equal to X2 + 6. So that means that DU is going to be equal to 2 X. DX. So making this substitution, we see that our integral here is going to be equal to the integral of minus 3, multiplied by 12 DU. Divided by You And so, perform the integration. This is going to be equal to. We can bring the constants out front, so we'll have minus 3 divided by 2, and that gets multiplied by the interval of 1 divided by U DU, which when we integrate that is going to give us the natural log of the absolute value of U plus C. However, we need everything in terms of X, so we'll substitute back in X2 + 6 for you. And so this is going to be equal to minus 3 divided by 2 multiplied by the natural log of the absolute value of the quantity of X2 + 6 in quantity plus C. So now, combining these results, we see that our original integral, which is the integral. Of 6 multiplied by the quantity of X2 plus 9 in quantity, divided by the quantity of X, multiplied by the quantity of X2 + 6 in quantity. DX is going to be equal to. 9 multiplied by the natural log of the absolute value of X. Minus 3 divided by 2 multiplied by the natural log of the quantity of X2 + 6 in quantity plus C. So here we drop the absolute value sign for the second natural log function since X2 + 6 is always going to be a positive 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.50. ∫ 8(x² + 4)/[x(x² + 8)] dx

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23-64. Integration Evaluate the following integrals.
50. ∫ 8(x² + 4)/[x(x² + 8)] dx