/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 28

Evaluate the following integrals. $$\int \frac{16 x^{2}}{(x-6)(x+2)^{2}} d x$$

Short Answer

Expert verified
Question: Evaluate the following integral: $$\int\frac{16x^2}{(x-6)(x+2)^2}dx$$ Answer: The evaluated integral is given by: $$\int\frac{16x^2}{(x-6)(x+2)^2}dx=9\ln|x-6|+7\ln|x+2|-\frac{8}{x+2}+C$$

Step by step solution

01

Decompose the fraction into partial fractions

Let's decompose the given function into partial fractions: $$\frac{16 x^{2}}{(x-6)(x+2)^{2}}=\frac{A}{x-6}+\frac{B}{x+2}+\frac{C}{(x+2)^{2}}$$ Now, we need to find the values of A, B, and C. To do this, clear the denominator by multiplying both sides by \((x-6)(x+2)^2\): $$16x^2 = A(x+2)^2 + B(x-6)(x+2) + C(x-6)$$
02

Solve for A, B, and C

To find the values of A, B, and C, we can choose convenient values of x to cancel some of the terms on the right-hand side and simplify the equation. Choose \(x=6\): $$16(6)^2 = A(8)^2$$ Solving for A, we get \(A=9\). Choose \(x=-2\): $$16(-2)^2 = C(-8)$$ Solving for C, we get \(C=8\). Next, we need to find B. To do this, we can choose a convenient value for x or compare coefficients. Let's compare the coefficients of \(x^2\): $$16 = 9 + B$$ Solving for B, we get \(B=7\). Now we have the decomposed function: $$\frac{16x^2}{(x-6)(x+2)^2}= \frac{9}{x-6} + \frac{7}{x+2} + \frac{8}{(x+2)^2}$$
03

Integrate each term

Now we can integrate each term separately: $$\int\frac{16x^2}{(x-6)(x+2)^2}dx = \int\frac{9}{x-6}dx + \int\frac{7}{x+2}dx + \int\frac{8}{(x+2)^2}dx$$ The first two integrations are simple natural logarithmic functions, and the third one is an inverse function. Let's find the integrations for each term: Integration of the first term: $$\int\frac{9}{x-6}dx = 9\int\frac{1}{x-6}dx = 9 \ln|x-6|+C_1$$ Integration of the second term: $$\int\frac{7}{x+2}dx = 7\int\frac{1}{x+2}dx = 7 \ln|x+2|+C_2$$ Integration of the third term: $$\int\frac{8}{(x+2)^2}dx = 8\int\frac{1}{(x+2)^2}dx=-\frac{8}{x+2}+C_3$$
04

Combine the integrations

Now, add up the three integrations to get the final result: $$\int\frac{16x^2}{(x-6)(x+2)^2}dx=9\ln|x-6|+7\ln|x+2|-\frac{8}{x+2}+C$$ where \(C = C_1+C_2+C_3\) is the constant of integration.

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