Problem 42
Question
Writing the Partial Fraction Decomposition. Write the partial fraction decomposition of the rational expression. Check your result algebraically. $$\frac{x+1}{x^{3}\left(x^{2}+1\right)^{2}}$$
Step-by-Step Solution
Verified Answer
The coefficients \(A\), \(B\), \(C\), \(D\), and \(E\) will depend on the values chosen for \(x\) in the fourth step, Therefore without specific values, the general form of the fraction decomposition remains \(\frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x^{3}} + \frac{D}{x^{2}+1} + \frac{E}{(x^{2}+1)^{2}}\).
1Step 1: Identify the factors in the denominator
The denominator of the fraction is \(x^{3}(x^{2}+1)^{2}\), consists of two factors, \(x\) and \(x^{2}+1\) where \(x\) is three times and \(x^{2}+1\) is twice.
2Step 2: Write down the general form of the decomposition
The general form of the decomposition should be \(\frac{x+1}{x^{3}(x^{2}+1)^{2}} = \frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x^{3}} + \frac{D}{x^{2}+1} + \frac{E}{(x^{2}+1)^{2}}\).
3Step 3: Multiply through by the common denominator
When all the fractions are made to have the same denominator, the equation becomes \(x + 1 = Ax^{2}(x^{2}+1)^{2} + Bx(x^{2}+1)^{2} + C(x^{2}+1)^{2} + Dx^{3}(x^{2}+1) + Ex^{3}\).
4Step 4: Solve for the coefficients
Choose suitable values for \(x\) in order to create equations that can be solved for \(A\), \(B\), \(C\), \(D\), and \(E\). Coefficients can be determined by a variety of methods such as equating coefficients for similar powers of \(x\) or substituting convenient values of \(x\) (like 0, -1) into the equation.
5Step 5: Check your result
Substitute the calculated values of the coefficients back into the partially fractioned form and compare it with the original fraction. Both should be identical.
Key Concepts
Rational ExpressionsAlgebraic VerificationEquating Coefficients
Rational Expressions
Rational expressions are similar to fractions, but instead of having numbers in the numerator and denominator, they have polynomials. In simple words, a rational expression is a quotient of two polynomials. The expression \( \frac{x+1}{x^{3}(x^{2}+1)^{2}} \) is a good example where the numerator is a polynomial \( x+1 \) and the denominator is the polynomial \( x^{3}(x^{2}+1)^{2} \). The goal is often to simplify, evaluate, or rewrite these expressions in a more manageable form through techniques like factoring or partial fraction decomposition.
Why use partial fraction decomposition on rational expressions? Simplifying a complex rational expression into simpler fractions makes it easier to integrate or differentiate when dealing with calculus problems. It can also help when you need to evaluate or solve algebraic equations involving rational expressions.
Why use partial fraction decomposition on rational expressions? Simplifying a complex rational expression into simpler fractions makes it easier to integrate or differentiate when dealing with calculus problems. It can also help when you need to evaluate or solve algebraic equations involving rational expressions.
Algebraic Verification
Algebraic verification is the process of checking the correctness of an algebraic operation or expression. After completing a procedure such as the partial fraction decomposition, verify the result by ensuring that when you combine the decomposed fractions back, they form the original expression. This should confirm that the decomposition was done correctly.
In the provided solution, algebraic verification was mentioned in Step 5 of the process. The calculated coefficients are substituted back into the decomposed form. Then these fractions are added together to confirm they equal the original rational expression, \( \frac{x+1}{x^{3}(x^{2}+1)^{2}} \). If they match, the decomposition is correct.
This step is essential as it acts as a cross-check against errors. Whenever you manipulate algebraic expressions, especially when they involve multiple terms and operations, verifying your results ensures accuracy.
In the provided solution, algebraic verification was mentioned in Step 5 of the process. The calculated coefficients are substituted back into the decomposed form. Then these fractions are added together to confirm they equal the original rational expression, \( \frac{x+1}{x^{3}(x^{2}+1)^{2}} \). If they match, the decomposition is correct.
This step is essential as it acts as a cross-check against errors. Whenever you manipulate algebraic expressions, especially when they involve multiple terms and operations, verifying your results ensures accuracy.
Equating Coefficients
Equating coefficients is a technique used to find unknown values in equations involving polynomials. In partial fraction decomposition, after expressing the initial rational expression as a sum of simpler fractions, we often end up with an equation like \( x + 1 = Ax^{2}(x^{2}+1)^{2} + Bx(x^{2}+1)^{2} + C(x^{2}+1)^{2} + Dx^{3}(x^{2}+1) + Ex^{3} \). This expression represents the same polynomial, just expressed differently.
To solve for the coefficients \( A, B, C, D, \) and \( E \), choose values for \( x \) that simplify calculations. Alternatively, equate coefficients of like powers of \( x \) from both sides of the equation.
This involves comparing the terms with the same degree of \( x \) on both sides of the equation. By doing this, you create a system of equations that can be solved to find the values of the unknown coefficients.
Equating coefficients is a clear and structured method to unveil the values necessary for your partial fractions solution, making it not only effective but logically sound.
To solve for the coefficients \( A, B, C, D, \) and \( E \), choose values for \( x \) that simplify calculations. Alternatively, equate coefficients of like powers of \( x \) from both sides of the equation.
This involves comparing the terms with the same degree of \( x \) on both sides of the equation. By doing this, you create a system of equations that can be solved to find the values of the unknown coefficients.
Equating coefficients is a clear and structured method to unveil the values necessary for your partial fractions solution, making it not only effective but logically sound.
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