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.

46. ∫ (x³ + 4x² + 12x + 4)/((x² + 4x + 10)²) dx

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to evaluate the interval, and we're given the interval of the quantity of XQ plus 6 X2 + 17 X plus 10 in quantity divided by the quantity of X2 + 6 X plus 13 in quantity squared X. The first step is to divide one factor of our denominator into our numerator. So, we're going to take X2 + 6 X + 13, and divide it into. Execute Plus 6 X squared. Plus 17 X. Plus 10 So, X2 will go into execute x times, and we'll Take X and multiplied by the quantity of X2 + 6 X + 13. In doing so, we will get X cubed. Plus 6 X squared. Plus 13 X. And we'll subtract that from our numerator. In doing so, we'll get as a remainder of 4 X plus 10. So that means that our integral of X cubed plus 6 X2 + 17 X. Plus 10, and that quantity is going to be divided by the quantity of X2 plus 6 X plus 13 in quantity squared DX. is going to be equal to the interval. Of X divided by the quantity of X2 + 6 X plus 13 in quantity DX. Plus The interval of the quantity of 4 X plus 10 in quantity, divided by the quantity of X2 + 6 X. Plus 13 in quantity squared DX. So we're gonna look at each one of these integrals separately. So, for the first integral, we have the integral, and here we're gonna rewrite this slightly. So we'll have the interval of X divided by the quantity of the quantity of X plus 3 and quantity squared. Plus 4 in quantity DX. So here we just completed the square in the denominator. So, we can integrate this using substitutions. We'll set U equal to X + 3, and that means that DU is going to be equal to the X, but we can also write X as being equal to U minus 3. So making this substitution, we see that this is going to be equal to. The integral of quantity of u minus 3 in quantity divided by the quantity of U2 + 4 in quantity D U. Now recall that the integral of U, divided by the quantity of U2 + 4 and quantity the U is going to be equal to. 1/2 multiplied by the natural log. Of the quantity of U2 + 4 in quantity plus D. and also recall that the interval. Of 1 divided by the quantity of U2 plus 4 in quantity DU is going to be equal to 12 multiplied by the inverse tangent of the quantity of U divided by 2 in quantity plus D. So, using those two results, we see that our integral here is going to be equal to. One half multiplied by the natural log of the quantity, and herefore you will substitute in back x + 3, and so that means that the argument of our natural log function is just going to be our original denominator, which is going to be X2 + 6 X + 13 in quantity, and this then for the next term we'll have minus, and here we have 3. Divided by 2. Multiplied by the inverse tangent of the quantity of X + 3 in quantity divided by 2 in quantity plus D. So here we just combined our two integration constants into a single integration constant. Next, we need to look at the next integral, which is going to be the integral. Of the quantity of 4 X. Plus 10 in quantity, divided by the quantity of X2 plus 6 X plus 13 in quantity squared. DX. Now, again, we would like to use substitution to evaluate this interval, so we'll set you equal to X2 + 6 X. Plus 13, and so that means that DU is going to be equal to the quantity of 2 X plus 6 in quantity DX. So in order to use a substitution, we'll need to rewrite our internal event. So that means our integral here is going to be equal to the integral. Of 2 multiplied by the quantity of 2 X plus 6 in quantity, divided by the quantity. of X2 + 6 X. Plus 13 in quantity squared. DX Minus The integral of 2 divided by. The quantity of X2 + 6 X. Plus 13 in quantity squared DX. So again, we need to evaluate two separate intervals, and we'll do these individually. For the first interval, we'll use our substitution and rewrite that as the interval. Of 2 divided by U squared. DU. And this is simple enough to integrate. So, recall that when we integrate 1 divided by U squared, we get minus 1 divided by U. So that means that this interval is going to be equal to -2 divided by U. Plus D. So next, we're gonna be looking at the interval. of 2 divided by The quantity, and here we're going to complete the square in the denominator again. So we'll have the quantity of X + 3. In quantity where Plus 2 squared in quantity squared X. And we call that the interval. Of one divided by. The quantity of U2 plus Aquan quantity squared the U. is going to be equal to. You Divided by the quantity Of 2 a squared. Multiplied by the quantity of U2 plus a squad in quantity. Plus, 1 divided by the quantity of 2 a cubed in quantity multiplied by the inverse tangent. Of the quantity of U divided by A in quantity plus C. So, in our case here, A is going to be equal to 2, and we use the substitution that U is equal to X plus 3, that means that DU is going to be equal to DX. So, our integral here is going to be equal to. The quantity of X + 3. Divided by the quantity, and here we have a factor of 2 in the numerator that we had from our original integral, and that's going to cancel out the factor of 2 in the denominator of each of our two terms. So, in our denominator here, we'll just have a squared, which is going to be 4, multiplied by the quantity, and here we're going to have our original denominator, which is going to be X2 + 6 X + 13. In quantity, then we'll have. Los 1 divided by 8, multiplied by the inverse tangent. Of the quantity of X + 3 in quantity. Divided by 2. In quantity plus D. So that means that the interval. Of the quantity of 4 X. Plus 10 in quantity, divided by the quantity of X squad, plus 6 X plus 13 in quantity squared DX is going to be equal to. -2 divided by, and here we'll substitute back in you into the result for our first integra, which was. The quantity of X2 plus 6 X. Plus 13 in quantity. And then we'll have minus. The quantity of X plus 3 in quantity, divided by the quantity of 4, multiplied by the quantity of X2 + 6 X. Plus 13 in quantity. -1 divided by 8, multiplied by the inverse tangent. Of the quantity of X + 3 in quantity, divided by 2 in quantity plus D. And we can actually combine those 1st 1st 2 terms, and that means that this integral is going to be equal to minus the quantity of X plus 11 in quantity, divided by the quantity of 4, multiplied by the quantity of X2 + 6 X + 13 in quantity, minus 1 divided by 8, multiplied by the inverse tangent. Of the quantity of X + 3 in quantity, divided by 2, in quantity plus D. And so now we just need to put all of our results together. So, the integral of the quantity of execute. Plus 6 X squared. Plus 17 X. Plus 10 in quantity, divided by the quantity of X2 + 6 X plus 13 in quantity squared DX is going to be equal to. One half multiplied by the natural log of quantity of X2 + 6 X + 13. In quantity Minus 3 divided by 2, multiplied by the inverse tangent. Of the quantity of X + 3 in quantity divided by 2 in quantity. Minus the quantity of X + 11. In quantity, divided by the quantity of 4 multiplied by the quantity of X squared. Plus 6 X plus 13 in quantity. 1 divided by 8, multiplied by the inverse tangent. Of the quantity of X + 3, in quantity divided by 2, in quantity plus D. Now, we can combine our two arc tangent terms, and so that means our final answer is going to be equal to. One half multiplied by the natural log of quantity of X2 + 6 X + 13 in quantity, minus. 13 Divided by 8, multiplied by the inverse tangent. Of the quantity of X plus 3 in quantity divided by 2 in quantity, minus. The quantity of X plus 11 and quantity divided by the quantity of 4 multiplied by the quantity of X2 + 6 X. Plus 13 in quantity. Plus D. So this here is the final 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.
46. ∫ (x³ + 4x² + 12x + 4)/((x² + 4x + 10)²) dx

Key Concepts

Polynomial Long Division and Simplification

Integration by Partial Fractions

Integration Techniques for Quadratic Denominators

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