Problem 25
Question
Use partial fractions to find the indefinite integral. $$ \int \frac{x^{2}-4 x-4}{x^{3}-4 x} d x $$
Step-by-Step Solution
Verified Answer
The indefinite integral of the given function is \(-2\ln|x| - \ln|x-2| + 2\ln|x+2| + C\).
1Step 1: Perform Polynomial Division
First, we need to check if the degree of the numerator \(x^{2}-4 x-4\) is greater than the degree of the denominator \(x^{3}-4 x\). In this case, the degree of the denominator is greater, so, we don't need to perform polynomial division.
2Step 2: Decompose into Partial Fraction
Next, we can factorize \(x^{3}-4 x\) into \(x(x^{2}-4)\) or \(x(x-2)(x+2)\). Our fraction becomes: \[\frac{x^{2}-4 x-4}{x(x-2)(x+2)}\]. Then we decompose the fraction into partial fractions. We write it in the form \[\frac{A}{x} + \frac{B}{x-2} + \frac{C}{x+2}\] where A, B, and C are constants to be found.
3Step 3: Find Constants A, B, and C
Multiply both sides by the denominator \(x(x-2)(x+2)\) to clear the fractions. Now solve for A, B, and C, we get: \(A = -2, B = -1, C = 2\). Therefore, the fraction decomposed into partial fractions is \[= -\frac{2}{x} - \frac{1}{x-2} + \frac{2}{x+2}\].
4Step 4: Compute Integral
The next step is to integrate our decomposed fraction. By integrating each term separately, the integral becomes \[-2\int\frac{dx}{x} - \int\frac{dx}{x-2} + 2\int\frac{dx}{x+2}\] which simplifies to \[-2\ln|x| - \ln|x-2| + 2\ln|x+2| + C\]. C is the constant of integration.
Other exercises in this chapter
Problem 25
Determine whether the improper integral diverges or converges. Evaluate the integral if it converges, and check your results with the results obtained by using
View solution Problem 25
Present Value In Exercises 25 and 26, use a program similar to the Simpson's Rule program on page 906 with \(n=8\) to approximate the present value of the incom
View solution Problem 25
Find the indefinite integral. (Hint: Integration by parts is not required for all the integrals.) $$ \int \frac{e^{1 / t}}{t^{2}} d t $$
View solution Problem 26
Determine whether the improper integral diverges or converges. Evaluate the integral if it converges, and check your results with the results obtained by using
View solution