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.

63. ∫ dx/(x² - 2x - 15)

Accepted Answer

Welcome back, everyone. Determine the following integral. Integral of D X divided by X2 + 5 X + 6. For this problem we're going to use partial fraction decomposition. Let's rewrite our integral. Our function is 1 divided by X2 + 5 X + 6. And we can notice that the quadratic expression is reducible. We can specifically factor it out in its linear factors. So we get 1 divided by X + 2 multiplied by X + 3. This is the factored form of our quadratic polynomial. Let's go ahead and rewrite our resultant fraction as a divided by the first linear factor X + 2. Plus B divided by our second linear factor X + 3. Now we can find the common denominator. We already know that it is X + 2. Multiplied by x + 3. Meaning A is going to be multiplied by X + 3. So we get A multiplied by X + 3 + B, which is multiplied by X + 2. And we can now show that A multiplied by X + 3 + B multiplied by X + 2 is going to be equal to 1, which is our numerator. Now we can expand this expression that would be AX plus 3A. Plus BX + 2B is equal to 1. And combining our X terms, we get A plus B, multiplied by X + 3 A plus 2B is equal to 1. We can now set a system of equations and show that A plus b must be equal to 0. That's because we don't have any X terms in the numerator. And then 3 A plus 2B must be equal to 1, because that's our free coefficient. From the first equation, we can show that B is equal to negative a. And substituting into the second expression we get 3 A plus 2 multiplied by negative A is equal to 1, meaning A is equal to 1 and thereforeb, which is negative a, must be -1. So we can now show that our integral becomes a divided by X + 2, which is 1 divided by X + 2, plus B divided by X + 3. So we're subtracting 1 divided by X + 3, and we're integrating with respect to X. Well then, and now we can integrate. We have basic integrals. The integral of D X divided by X + 2 is LN of the absolute value of X + 2. And we're subtracting the integral of D X divided by X + 3. The integral would be LN of the absolute value of X + 3. We're going to add a constant of integration C for an indefinite integral, and now using the properties of logarithms, we can rewrite this answer as a line of the absolute value of X + 2. Divided by X + 3. Plus a constant of integration C. That would be our final answer and thank you for watching.

2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.63. ∫ dx/(x² - 2x - 15)

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2–74. Integration techniques Use the methods introduced in Sections 8.1 through 8.5 to evaluate the following integrals.
63. ∫ dx/(x² - 2x - 15)