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Resolve into partial fractions $\frac{x^{2}+1}{x(x^{2}-1)}$

written 5.2 years ago by teamques10 ★ 68k

modified 2.6 years ago by pedsangini276 • 4.8k

engineering mathematics

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written 5.2 years ago by teamques10 ★ 68k

Solution:

$\frac{x^{2}+1}{x(x+1)(x-1)} = \frac{A}{x} +\frac{B}{x+1} + \frac{c}{x-1}$

$x^{2}+1 = A(x-1)(x+1)+B(x)(x-1)+C(x)(x+1)$

put x = 0

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

A = -1

put x = -1

$(-1)^{2} + 1 = B(-1)(-1-1)$

B = 1

put x = 1

$1^{2} + 1 = C(1)(1+1)$

C = 1

$\frac{x^{2}+1}{x(x+1)(x-1)} = \frac{-1}{x} +\frac{1}{x+1} + \frac{1}{x-1}$

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