23-64. Integration Evaluate the following integrals.
57....

Question

23-64. Integration Evaluate the following integrals.

57. ∫ (x³ + 5x)/(x² + 3)² 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 the quantity of XQ plus 2 X in quantity, divided by the quantity of X2 + 1 in quantity squared DX. So the first thing we want to do is to rewrite the fraction that is our integram into multiple fractions using partial fractions. So we call for partial fractions that the quantity of X cubed plus 2 X in quantity, divided by the quantity of X2 + 1 in quantity squared is going to be equal to the quantity of AX plus B in quantity, divided by the quantity of X2 + 1. In quantity, plus the quantity of DX plus D in quantity, divided by the quantity of X2 + 1 in quantity squared. So the next step is to multiply both sides of the equation by the quantity of X2 + 1 in quantity squad. In doing so, we will see that X cubed plus 2 X is equal to the quantity of AX plus B. In quantity, multiplied by the quantity of X2 + 1 in quantity, plus D X plus D. So multiplying everything out on the right hand side, we will get X cubed plus 2 X is equal to. A execute. Plus A X. Plus, B X squared. Plus B. Plus the X. Plus D. and collecting like terms, we'll see that X cubed plus 2 X. is equal to A execute. Plus BX2. Plus the quantity of A plus C in quantity, multiplied by X. Plus B plus D. Now we need to do is to match coefficients with like powers of X on both sides of the equation, and if we look at the cubic term, we'll see that A is going to be equal to 1. If we look at the quadratic term, we'll see that B is equal to 0, since we have no quadratic term on the left hand side of our equation. For the linear term, we'll see that A plus D is equal to 2, and for the constant term, we will have that B plus D is equal to 0. So since from the first equation, A is equal to 1, we can use that value for A and substitute into the third equation to solve for C. In doing so, we'll get that 1 plus D is equal to 2, and solving for C, we'll see that C is going to be equal to 1. And we can also take the second equation, which says B is equal to 2, and plug that value of B into the. 4th equation here, in doing so, we can solve for D, and so we'll have 0 plus D is equal to 0, and that means that D is going to be equal to 0. So now we have found all four of those values. We can now rewrite our integrals. So our original integral, which was the integral of the quantity of X cubed plus 2 X in quantity, divided by the quantity of X2 + 1 in quantity squared DX. is going to be equal to the integral. of X, divided by the quantity of X2 + 1 in quantity, D X. Plus the interval of X divided by the quantity of X2 + 1 in quantity squared DX. And now to integrate both of these integrals, we'll have to make a substitution, and we'll be making the same substitution for both integrals. So the substitution that we want to make is to set U equal to X2 + 1, and that means that DU is going to be equal to 2 X. DX. But if we look at our ingram, we see that we have just XDX. And so we'll divide both sides of that equation by 2, and so we'll have 1/2 multiplied by DU it's going to be equal to X multiplied by DX. So making this substitution, our integrals are going to be equal to the integral of 1 divided by U, multiplied by 1/2. Do you? Plus the interval of 1 divided by U squared multiplied by 1/2. Do you And so now we can perform the integration. So, for the first integral, it's going to be equal to. We'll have the constant of one half that we can pull out front, so we'll have 1/2, and that gets multiplied by the integral of 1 divided by U, DU, and recall that when we integrate 1 divided by U with respect to you, that we get the natural log of the absolute value of you. And then we'll have plus for the second integral, we'll also factor out the 1/2 in front, since it's constant, so we'll have 1/2, then we'll be multiplying this by the integral of 1 divided by U squared with respect to you. And recall that when we perform that integration that we get minus 1 divided by U. And then we'll have plus C since we'll have an integration constant from both of these integrals, which we can combine into a single integration constant of plus C. However, we want our final answer in terms of X instead of you, so we'll substitute back in X2 + 1 for you. In doing so, this is going to be equal to 1/2 multiplied by the natural log of the absolute value of the quantity of X2 + 1 in quantity, minus 1 divided by the quantity of 2 multiplied by the quantity of X2 + 1 in quantity. Plus C. And simplifying further, this is going to be equal to 1/2 multiplied by the natural log of the quantity of X2 + 1 in quantity minus 1 divided by the quantity of 2 X2 + 2 in quantity plus C, where we have dropped the absolute value sign inside the natural log function, since X2 + 1 is always going to be a positive quantity. And this answer here is the answer that we're looking for for this problem. So, I hope this explanation has been useful, and I'll see you in the next video.

23-64. Integration Evaluate the following integrals.57. ∫ (x³ + 5x)/(x² + 3)² dx

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23-64. Integration Evaluate the following integrals.
57. ∫ (x³ + 5x)/(x² + 3)² dx