23-64. Integration Evaluate the following integrals.
54....

Question

23-64. Integration Evaluate the following integrals.

54. ∫ (z + 1)/[z(z² + 4)] dz

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says determine the following integral, and we're getting the integral of the quantity of Y + 2 in quantity, divided by the quantity of Y multiplied by the quantity of Y2 + 9 in quantity D Y. So the first step in evaluating this interval is to take our integrain here and break it up into multiple fractions using partial fractions. If we call for partial fractions that the quantity of Y plus 2 in quantity, divided by the quantity of Y multiplied by the quantity of Y squared plus 9 in quantity is going to be equal to a divided by Y. Plus the quantity of B Y plus D and quantity divided by the quantity of Y2 plus 9. The next step is to multiply both sides of this equation by the denominator on the left hand side. In doing so, we'll get that Y + 2 is equal to a multiplied by the quantity of Y2 + 9 in quantity. Plus the quantity of BY plus C in quantity multiplied by Y. Multiplying everything out on the right hand side, we'll get that Y + 2 is going to be equal to A Y2 + 9 A. Plus, By squared. Plus the Y. and collecting like terms, we'll see that Y + 2 is going to be equal to the quantity of A plus B in quantity multiplied by Y squared. Plus DY + 9 A. And so here we need to match the coefficients of the various powers Y on both sides of the equation. So, for the quadratic term, we see that A plus B is going to be equal to 0, since we have no quadratic term on the left-hand side of the equation. For the linear terms, we'll see that we have C is equal to 1, and for the constant term, we'll have 9 A is equal to 2. Now, our second equation here gives us the value for C that C is equal to 1. And our last equation that we wrote, We can use to find the value of A. So taking that last equation and dividing both sides of that equation by 9, we'll get that A is equal to 2 divided by 9. We can use this value in the first equation to solve for B. So we'll have 2 divided by 9, plus B is equal to 0, and subtracting 2 divided by 9 for both sides of the equation, we'll get that B is equal to minus 2 divided by 9. So that means that the integral of the quantity of Y plus 2 in quantity divided by the quantity of Y multiplied by the quantity of Y quad plus 9 in quantityd Y is going to be equal to the interval of the quantity of 2 divided by 9 in quantity divided by Y D Y. Plus, the integral of the quantity of minus 2 divided by 9 multiplied by Y plus 1 in quantity, divided by the quantity of Y squared, plus 9 in quantity D Y. So what we want to do is to evaluate each one of these integrals independently. So looking at the first integral, we'll have the integral of the quantity of 2 divided by 9 in quantity, divided by Y D Y. So this is going to be equal to. We can take our constant and pull it in front of the integral. So we'll have 2 divided by 9, and this gets multiplied by, and here we're going to be integrating 1 divided by Y with respect to Y. And recall that that will give us the natural log of the absolute value of Y. And then we'll have to add our integration constant of plus C to this expression. Now for the 2nd interval, we will have the interval. Of the quantity of minus 2 divided by 9 multiplied by Y plus 1 in quantity, divided by the quantity of Y quad plus 9 in quantity Dy. And we'll want to separate this up into two separate integrals, but this is going to be equal to the integral of the quantity of minus 2 divided by 9 multiplied by Y in quantity, divided by the quantity of Y squared. Plus 9 in quantity DY. Plus the integral of 1 divided by the quantity of Y squared plus 9 in quantity D Y. Now, to evaluate that first integral, we're gonna need to make a substitution. So we're gonna set U equal to Y2 + 9, and that means that DU is going to be equal to 2 Y D Y. So, for that substitution, our expression here is going to be equal to. Minus 1 divided by 9 multiplied by the interval of 1 divided by U DU, then we'll have plus for the second integral we can integrate this directly. So recall that we integrate 1 divided by the quantity of Y2 + 9 in quantity with respect to Y, that we will get 1 divided by 3 multiplied by the inverse tangent of the quantity of Y divided by 3. In quantity plus C or C here is our integration constant. So now we just need to evaluate the other interval. In doing so, this is going to be equal to. -1 divided by 9, multiplied by, and here when we integrate. One divided by you with respect to you, that's going to give us the natural log of the absolute value of you. Then we'll have plus 1 divided by 3, multiplied by the inverse tangent of the quantity of Y divided by 3 in quantity plus D. Well, we've taken both integration constants here and and combined them into a single integration constant. However, we want our final answer in terms of Y. So for you, we'll substitute back in Y2 + 9. So, in doing so, this is going to be equal to minus 1 divided by 9, multiplied by the natural log of the absolute value of the quantity of Y2 plus 9 in quantity, plus 1 divided by 3 multiplied by the inverse tangent. Of the quantity of Y divided by 3 in quantity plus Z. So now we just need to combine everything together. And so the integral of the quantity of Y plus 2 in quantity, divided by the quantity of Y multiplied by the quantity of Y squared plus 9 in quantity DY is going to be equal to. 2 divided by 9, multiplied by the natural log of the absolute value of Y. Minus 1 divided by 9, multiplied by the natural log of the quantity of Y2 + 9 in quantity. Plus 1 divided by 3, multiplied by the inverse tangent. Of the quantity of Y divided by 3 in quantity plus D. So here we combine all of our integration concepts into a single integration concept of plus C. And also in the second natural log term, we remove the absolute value um signs because Y2 + 9 is always going to be a positive quantity. So now we want to simplify this expression. So, we can actually use the properties of logarithms to rewrite our first term, to remove the absolute value sign. So this is going to be equal to 1 divided by 9, multiplied by the natural log of the quantity of Y2 in quantity, minus 1 divided by 9, multiplied by the natural log of the quantity of Y2 plus 9 in quantity. Plus 1 divided by 3, multiplied by the inverse tangent of the quantity of Y divided by 3 in quantity plus C. So here we've used property logarithms to rewrite two multiplied by the natural log of the absolute value Y as just the natural log of the quantity of Y squared in quantity. Since Y2d is a positive quantity, we can drop the absolute value sign. And so now we can combine our two natural log terms. In doing so, this is going to be equal to 1 divided by 9, multiplied by the natural log of the quantity of Y2d divided by the quantity of Y2. Plus 9 in quantity, plus 1 divided by 3, multiplied by the inverse tangent. Of the quantity of Y divided by 3 in quantity plus D. 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.
54. ∫ (z + 1)/[z(z² + 4)] dz

Key Concepts

Partial Fraction Decomposition

Integration of Rational Functions

Handling Quadratic Denominators

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