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Magazine Chapter 8: Problem 19
Find the integral. $$ \int \frac{-x}{x^{3}-3 x^{2}+2 x} d x $$
Short Answer
Expert verified
The integral is \( \ln \left| \frac{x-1}{x-2} \right| + C \).
Step by step solution
01
Simplify the Integrand
Factor the denominator \( x^3 - 3x^2 + 2x \). First, factor out an \( x \) to get \( x(x^2 - 3x + 2) \). Next, factor the quadratic \( x^2 - 3x + 2 \) into \( (x-1)(x-2) \). Now, the expression becomes \( \frac{-x}{x(x-1)(x-2)} \).
02
Set Up Partial Fraction Decomposition
Express \( \frac{-x}{x(x-1)(x-2)} \) as a sum of partial fractions: \( \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x-2} \).
03
Combine Partial Fractions
Multiply both sides by the common denominator \( x(x-1)(x-2) \) to obtain the equation:\(-x = A(x-1)(x-2) + Bx(x-2) + Cx(x-1)\).
04
Expand and Collect Terms
Expand the right side:- \( A(x-1)(x-2) = A(x^2 - 3x + 2) \)- \( Bx(x-2) = B(x^2 - 2x) \)- \( Cx(x-1) = C(x^2 - x) \)Combine like terms to form the equation:\(-x = (A + B + C)x^2 + (-3A - 2B - C)x + 2A \).
05
Solve for Coefficients
Equate the coefficients from both sides of the equation to get the following system of equations:- \( A + B + C = 0 \)- \( -3A - 2B - C = -1 \)- \( 2A = 0 \)From \( 2A = 0 \), we find \( A = 0 \). Substituting \( A = 0 \) into the other equations gives \( B + C = 0 \) and \( -2B - C = -1 \). Solving these, \( B = 1 \) and \( C = -1 \).
06
Write the Integral in Terms of Partial Fractions
The decomposition becomes \( \frac{1}{x-1} - \frac{1}{x-2} \). The integral becomes: \[ \int \left( \frac{1}{x-1} - \frac{1}{x-2} \right) dx \].
07
Integrate Each Term
Integrate each term separately:- \( \int \frac{1}{x-1} \, dx = \ln|x-1| + C_1 \)- \( \int \frac{1}{x-2} \, dx = \ln|x-2| + C_2 \)Combining these results, we have:\[ \ln|x-1| - \ln|x-2| + C \].
08
Simplify the Result
Use the properties of logarithms to combine the result into a single logarithm:\[ \ln \left| \frac{x-1}{x-2} \right| + C \]. This is the final formula for the integral.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Fraction Decomposition
Partial fraction decomposition is a technique used in integral calculus to break down complex rational expressions into simpler fractions that are easier to integrate. When you have an integrand that is a rational expression, like our given expression \( \frac{-x}{x^{3}-3 x^{2}+2 x} \), where the degree of the numerator is less than the degree of the denominator, partial fraction decomposition is a handy tool.
- First, we express the integrand as a sum of simpler fractions.
- This involves finding constants, such as \( A \), \( B \), and \( C \), which turn complex expressions into easily integrable ones.
Factoring Polynomials
Factoring polynomials is a crucial preparatory step before applying partial fraction decomposition. In this exercise, we factor the denominator \( x^3 - 3x^2 + 2x \).
- Begin by factoring out the greatest common factor. In our case, that's \( x \), yielding \( x(x^2 - 3x + 2) \).
- Next, we factor the quadratic \( x^2 - 3x + 2 \). Using the factoring method for quadratics, we find it factors to \( (x-1)(x-2) \).
Definite and Indefinite Integrals
Integrals in calculus can be either definite or indefinite, and understanding their difference is vital. This exercise focuses on indefinite integrals.
- An indefinite integral, also known as an antiderivative, is represented by the integral symbol with no upper or lower limits. It often includes a constant \( C \) because antiderivatives do not have a unique solution without additional information.
- A definite integral calculates the accumulated value, such as area under a curve, between two specified bounds, and thus results in a numerical value rather than a function.
