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Partial fraction decomposition?
Use partial fraction expansion on
[-1] / [(x+1)(x^2+1)]
to get:
-1/2 1/(x+1) + 1/2 x/(x^2+1) -1/2 1/(x^2+1)
step by step please
2 Answers
- cidyahLv 73 years ago
-1/((x+1)(x^2+1)) = A/(x+1) +(Bx+C)/(x^2+1)
Multiply both sides by (x+1)(x^2+1)
-1 = A(x^2+1) + (Bx+C)(x+1)
Let x=-1
-1 = A((-1)^2+1) + 0
-1 = A(2)
2A = -1
A = -1/2
Compare the coefficients of x^2
0 = A+B
B = -A
B = 1/2
Compare the coefficients of x
0 = B + C
C = -B
C = -1/2
A=-1/2; B=1/2; C=-1/2
-1/((x+1)(x^2+1)) = A/(x+1) +(Bx+C)/(x^2+1)
-1/((x+1)(x^2+1)) = (-1/2)(1/(x+1)) + (1/2) x /(x^2+1) - (1/2) (1/(x^2+1))
= -1 /(2(x+1)) + x /(2(x^2+1)) - 1 /(2(x^2+1))
= -1 /(2x+2) + x/(2x^2+2) - 1 /(2x^2+2)
- 3 years ago
a/(x + 1) + bx/(x^2 + 1) + c/(x^2 + 1) = -1 / ((x + 1) * (x^2 + 1))
(a * (x^2 + 1) + bx * (x + 1) + c * (x + 1)) / ((x + 1) * (x^2 + 1)) = -1 / ((x + 1) (x^2 + 1))
a * (x^2 + 1) + b * (x^2 + x) + c * (x + 1) = 0x^2 + 0x - 1
ax^2 + bx^2 + bx + cx + a + c = 0x^2 + 0x - 1
a + b = 0
b + c = 0
a + c = -1
a = -b
c = -b
-b + -b = -1
-2b = -1
b = 1/2
a = -1/2
c = -1/2
(-1/2) / (x + 1) + (1/2) * x / (x^2 + 1) - (1/2) / (x^2 + 1)
