$\frac{x^2+1}{x^2(x+3)}$
Answer (1)
Set up the partial fraction decomposition: x 2 ( x + 3 ) x 2 + 1 = x A + x 2 B + x + 3 C .
Clear denominators and expand: x 2 + 1 = ( A + C ) x 2 + ( 3 A + B ) x + 3 B .
Solve for the coefficients: A = − 9 1 , B = 3 1 , C = 9 10 .
Write the final decomposition: x 2 ( x + 3 ) x 2 + 1 = − 9 x 1 + 3 x 2 1 + 9 ( x + 3 ) 10 .
Explanation
Understanding the Problem We are given a rational function x 2 ( x + 3 ) x 2 + 1 and asked to perform partial fraction decomposition.
Analyzing the Denominator The denominator is x 2 ( x + 3 ) , which has a repeated linear factor x 2 and a distinct linear factor ( x + 3 ) .
Setting up the Decomposition We set up the partial fraction decomposition as x 2 ( x + 3 ) x 2 + 1 = x A + x 2 B + x + 3 C , where A, B, and C are constants to be determined.
Clearing Denominators Multiply both sides of the equation by x 2 ( x + 3 ) to clear the denominators: x 2 + 1 = A x ( x + 3 ) + B ( x + 3 ) + C x 2 .
Expanding the Equation Expand the right side of the equation: x 2 + 1 = A x 2 + 3 A x + B x + 3 B + C x 2 .
Grouping Like Terms Group like terms: x 2 + 1 = ( A + C ) x 2 + ( 3 A + B ) x + 3 B .
Equating Coefficients Equate the coefficients of the corresponding powers of x on both sides of the equation to form a system of linear equations:
Coefficient of x 2 : A + C = 1 .
Coefficient of x : 3 A + B = 0 .
Constant term: 3 B = 1 .
Solving for B Solve the system of linear equations for A, B, and C.
From 3 B = 1 , we get B = 3 1 .
Solving for A Substitute B = 3 1 into 3 A + B = 0 to get 3 A + 3 1 = 0 , so 3 A = − 3 1 , and A = − 9 1 .
Solving for C Substitute A = − 9 1 into A + C = 1 to get − 9 1 + C = 1 , so C = 1 + 9 1 = 9 10 .
Substituting Values Substitute the values of A, B, and C back into the partial fraction decomposition: x 2 ( x + 3 ) x 2 + 1 = x − 9 1 + x 2 3 1 + x + 3 9 10 .
Simplifying the Expression Simplify the expression: x 2 ( x + 3 ) x 2 + 1 = − 9 x 1 + 3 x 2 1 + 9 ( x + 3 ) 10 .
Final Answer Thus, the partial fraction decomposition is: x 2 ( x + 3 ) x 2 + 1 = − 9 x 1 + 3 x 2 1 + 9 ( x + 3 ) 10
Examples
Partial fraction decomposition is used in calculus to integrate rational functions. It's also used in engineering to analyze systems and circuits, breaking down complex transfer functions into simpler components.
