Partial Fractions?

(x^2+x-1)/((x(x+1)^2)) = A/x + B/(x+1)+ C/(x+1)^2

multiply both sides by x(x+1)^2

x^2+x-1 = A(x+1)^2 + Bx(x+1) + Cx

Let x=-1

-1 = A(0) + B(0) + C(-1)

-C = -1

C = 1

Let x=0

-1 = A(1)^2

A = -1

Match the coefficient of x^2

1 = A + B

1 = -1 + B

B = 2

A=-1; B=2; C = 1

(x^2+x-1) /(x(x+1)^2) = -1/x + 2/(x+1) + 1/(x+1)^2

(x^2+x-1) /(x(x+1)^2) +1/x = -1/x + 2/(x+1) + 1/(x+1)^2 +1/x

(x^2+x-1) /(x(x+1)^2) +1/x = 2/(x+1) + 1/(x+1)^2

(x^2 + x - 1) / (x * (x + 1)^2) = a/x + b/(x + 1) + c/(x + 1)^2

a * (x + 1)^2 + b * x * (x + 1) + c * x = x^2 + x - 1

a * (x^2 + 2x + 1) + b * (x^2 + x) + cx = x^2 + x - 1

ax^2 + bx^2 + 2ax + bx + cx + a = x^2 + x - 1

a + b = 1

2a + b + c = 1

a = -1

-1 + b = 1

b = 2

2 * (-1) + 2 + c = 1

-2 + 2 + c = 1

c = 1

-1/x + 2/(x + 1) + 1/(x + 1)^2

Are you sure about that appended 1/x bit?

(x^2 + x - 1) / (x * (x + 1)^2) + 1/x =>

(x^2 + x - 1) / (x * (x + 1)^2) + (x + 1)^2 / (x * (x + 1)^2) =>

(x^2 + x - 1 + (x + 1)^2) / (x * (x + 1)^2) =>

(x^2 + x - 1 + x^2 + 2x + 1) / (x * (x + 1)^2) =>

(2x^2 + 3x) / (x * (x + 1)^2) =>

x * (2x + 3) / (x * (x + 1)^2) =>

(2x + 3) / (x + 1)^2

(2x + 3) / (x + 1)^2 = a/(x + 1) + b/(x + 1)^2

2x + 3 = a * (x + 1) + b

2x + 3 = ax + a + b

a = 2

a + b = 3

2 + b = 3

b = 1

(x^2 + x - 1) / (x * (x + 1)^2) + 1/x = 2/(x + 1) + 1/(x + 1)^2

(x² + x − 1) / [ x (x+1)² ] + 1/x

Use partial fraction decomposition on the first term. For repeated roots, remember to include all the powers.

A/x + B/(x+1) + C/(x+1)²

Recombine into one fraction:

[ A(x+1)² + Bx(x+1) + Cx ] / [ x (x+1)² ]

Focusing on the numerator:

A(x+1)² + Bx(x+1) + Cx

A(x²+2x+1) + Bx² + Bx + Cx

Ax² + 2Ax + A + Bx² + Bx + Cx

(A+B)x² + (2A+B+C)x + A

Match the coefficients to the original numerator (x² + x − 1):

A + B = 1

2A + B + C = 1

A = -1

Therefore, B = 2 and C = 1.

Substituting it all in:

A/x + B/(x+1) + C/(x+1)²

-1/x + 2/(x+1) + 1/(x+1)²

The -1/x will cancel out with the other term, so the final result is:

2/(x+1) + 1/(x+1)²

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