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Chapter 7: Problem 18

Evaluate the following integrals. $$\int \frac{21 x^{2}}{x^{3}-x^{2}-12 x} d x$$

Short Answer

Expert verified
Question: Evaluate the following integral $$\int \frac{21x^2}{x^3 - x^2 - 12x} dx$$ Answer: $$-\ln|x| + 12\ln|x - 4| - 3\ln|x + 3| + C$$

Step by step solution

01

Factor the denominator

Let's factor the denominator of the fraction: $$x^{3}-x^{2}-12x = x(x^2 - x - 12) = x(x - 4)(x + 3)$$
02

Perform partial fraction decomposition

Now, let's decompose the fraction into partial fractions: $$\frac{21x^2}{x(x - 4)(x + 3)} = \frac{A}{x} + \frac{B}{x-4} + \frac{C}{x+3}$$ Multiplying both sides with the denominator, we get: $$21x^2 = A(x - 4)(x + 3) + Bx(x + 3) + Cx(x - 4)$$
03

Solve for A, B, and C

To solve for A, B, and C, we can plug in x values that will make some terms cancel out. First, let x = 0: $$21(0)^2 = A(0 - 4)(0 + 3) \Rightarrow A=(-1)$$ Next, let x = 4: $$21(4)^2 = B(4)(4 + 3) \Rightarrow B=12$$ Finally, let x = -3: $$21(-3)^2 = C(-3)(-3 - 4) \Rightarrow C=(-3)$$ Now, we can rewrite the initial fraction as follows: $$\frac{-1}{x}+\frac{12}{x-4}+\frac{-3}{x+3}$$
04

Integrate each term

Now, let's evaluate the integral by integrating each term separately: $$\int \frac{-1}{x}+\frac{12}{x-4}+\frac{-3}{x+3} dx = -\int \frac{1}{x} dx + 12 \int \frac{1}{x - 4} dx - 3 \int \frac{1}{x + 3} dx$$ These integrals can be solved by recognizing the antiderivatives of the functions: $$-\ln|x| + 12\ln|x - 4| - 3\ln|x + 3| + C$$ This is the final expression for the integral. Don't forget to include the constant of integration (C).

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