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Chapter 7: Problem 22

Evaluate the following integrals. $$\int \frac{4 x-2}{x^{3}-x} d x$$

Short Answer

Expert verified
Question: Evaluate the integral $$\int\frac{4x-2}{x^3-x} dx$$. Answer: The integral evaluates to $$2\ln|x| + \ln|x-1| - \frac{3}{2}\ln|x+1| + C$$.

Step by step solution

01

Perform Partial Fraction Decomposition on the Integrand

First, we need to perform partial fraction decomposition on the integrand: $$\frac{4x-2}{x^3-x}$$ We can factor out the denominator to get: $$\frac{4x-2}{x(x^2-1)}$$ Now, we can factor the quadratic term in the denominator: $$\frac{4x-2}{x(x-1)(x+1)}$$ In partial fraction decomposition, we can write the integrand as the sum of three fractions: $$\frac{4x-2}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}$$ We can rewrite this as: $$(4x-2) = A(x-1)(x+1) + Bx(x+1) + Cx(x-1)$$
02

Determine Coefficients A, B, C

Now, we must find the values of A, B, and C. We can do this by choosing appropriate values of x to eliminate A, B, C one by one. Let's start: For A: Set \(x=0\), so the equation becomes: $$-2 = (-1)(1)A \Rightarrow A=2$$ For B: Set \(x=1\), so the equation becomes: $$2 = (1)(2)B \Rightarrow B=1$$ For C: Set \(x=-1\), so the equation becomes: $$-6 = (-2)(-2)C \Rightarrow C=-\frac{3}{2}$$ So, the partial fraction decomposition of our integrand is: $$\frac{2}{x} + \frac{1}{x-1} - \frac{3}{2(x+1)}$$
03

Integrate Each Term

Now that we have decomposed our integrand, we can integrate each term: $$\int\frac{4x-2}{x^3-x} dx = \int\left(\frac{2}{x} + \frac{1}{x-1} - \frac{3}{2(x+1)}\right) dx$$ Integrating each term, we get: $$2\int\frac{1}{x}dx + \int\frac{1}{x-1}dx - \frac{3}{2}\int\frac{1}{x+1}dx$$ Using the fact that \(\int\frac{1}{u}du = \ln|u|\) we get the result: $$2\ln|x| + \ln|x-1| - \frac{3}{2}\ln|x+1| + C$$ So, the final evaluated integral is: $$\int\frac{4x-2}{x^3-x} dx = 2\ln|x| + \ln|x-1| - \frac{3}{2}\ln|x+1| + C$$

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