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Chapter 7: Problem 27

Evaluate the following integrals. $$\int \frac{81}{x^{3}-9 x^{2}} d x$$

Short Answer

Expert verified
Based on the solution provided: Evaluate the integral $$\int \frac{81}{x^3-9x^2}dx$$. Answer: $$-9\ln|x - 9| + C$$

Step by step solution

01

Factor the denominator

We can factor out the common factor of \(x^2\) from the denominator: $$x^3 - 9x^2 = x^2(x - 9)$$ So, the integrand becomes: $$\frac{81}{x^2(x - 9)}$$
02

Use partial fraction decomposition

We rewrite the integrand as a sum of two simpler fractions: $$\frac{81}{x^2(x - 9)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 9}$$ Now, we will find constants A, B, and C. We multiply each side of the equation by the common denominator \(x^2(x - 9)\) to get rid of the denominators: $$81 = A x(x - 9) + Bx(x - 9) + Cx^2$$
03

Find the constants A, B, and C

Let's plug in some values for x to find A, B, and C: \(x = 0\): $$81 = 0 + 0 + (-9)C \implies C = -9$$ \(x = 9\): $$81 = 9A(0) + 9B(0) + 0 \implies$$ A and B don't affect this equation. Differentiate the equation with respect to x to obtain a simpler equation for finding A and B: $$0 = A(x - 9) + Ax + Bx$$ To find A, let \(x = 9\): $$0 = 9(0) + 9A \implies A = 0$$ To find B, differentiate the equation again with respect to x: $$0 = A + B$$ Now, we know that A = 0, so B = 0 as well.
04

Integrate the simpler fractions

Now, we rewrite the original integral with the values of A, B, and C that we found: $$\int \frac{81}{x^3-9x^2}dx = \int \left(\frac{0}{x} + \frac{0}{x^2} - \frac{9}{x-9}\right)dx$$ This integral simplifies to: $$-9 \int \frac{1}{x - 9} dx$$ Now, we can easily integrate this: $$-9 \int \frac{1}{x - 9} dx = -9 \ln|x - 9| + C$$ So, the final antiderivative is: $$-9\ln|x - 9| + C$$

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