Find the partial fraction decomposition for the following rational expression. (x^3+x^2+2x+7)/((x^2+1)^2)=(A)/(x^2+1)+(B)/((x^2+1)^2)

beatricalwu

beatricalwu

Answered question

2022-07-29

Find the partial fraction decomposition for the following rational expression.
x 3 + x 2 + 2 x + 7 ( x 2 + 1 ) 2 = A x 2 + 1 + B ( x 2 + 1 ) 2

Answer & Explanation

Steppkelk

Steppkelk

Beginner2022-07-30Added 11 answers

Following is the explanation for Partial Fraction Method.
1. Find if the Fraction is Proper or Improper.
If the degree of the numerator is less than the degree of the denominator then the fraction is proper
If the degree of the numerator is greater than or equal to the degree of the denominator then the fraction is improper
PROPER FRACTIONS
1. For Fractions that donot involve repeated factors the partial fractions can be represented by
1 x ( x 2 ) = A x + B x 2
There are two possibilities x and x-2 in the denominator
2. For Fractions that involve repeated factors the partial fractions can be represented by
1 x ( x 2 ) 2 = A x + B x 2 + B ( x 2 ) 2
There are three possibilities x, x-2 and ( x 2 ) 2 in the denominator
3. For Fractions that involve quadratic factors the partial fractions can be represented by
1 x ( x 2 + 2 ) = A x + B x + C x 2 + 2
There are two possibilities x and ( x 2 + 2 ) in the denominator and there is a possibility that the numerator in the second term contains the term of x
Since there is ( x 2 + 1 ) 2 term implies the partial fraction decomposition has two terms with ( x 2 + 1 ) and ( x 2 + 1 ) 2 in the denominator and the numerator is a linear term of the form Ax+B
Therefore, the Partial Fraction of decomposition can be written as
x 3 + x 2 + 2 x + 7 ( x 2 + 1 ) 2 = A x + B x 2 + 1 + C x + D ( x 2 + 1 ) 2
Multiply both sides by ( x 2 + 1 ) 2
x 3 + x 2 + 2 x + 7 = ( A x + B ) ( x 2 + 1 ) + ( C x + D ) x 3 + x 2 + 2 x + 7 = ( A x 3 + A x + B x 2 + B + C x + D )
Simplify
x 3 + x 2 + 2 x + 7 = A x 3 + B x 2 + ( A + C ) x + ( B + D )
Equate the coefficients on both sides
A = 1 , B = 1 , A + C = 2 C = 2 1 , B + D = 7 D = 7 1 = 6
Therefore, the partial fraction decomposition is
x 3 + x 2 + 2 x + 7 ( x 2 + 1 ) 2 = x + 1 x 2 + 1 + x + 6 ( x 2 + 1 ) 2

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