The partial fractions of x2+1xx2-1 are

x2+1x(x2-1)=Ax+Bx+1+Cx-1x2+1=A(x2-1)+Bx(x-1)+Cx(x+1)x=0 ⇒ A=-1x=1 ⇒ 2C=2 ⇒ C=1x=-1 ⇒ 2B=2 ⇒ B=1x2+1x(x2-1)=-1x+1x+1+1x-1

The partial fractions of x2+1xx2-1 are

x2+1x(x2-1)=Ax+Bx+1+Cx-1x2+1=A(x2-1)+Bx(x-1)+Cx(x+1)x=0 ⇒ A=-1x=1 ⇒ 2C=2 ⇒ C=1x=-1 ⇒ 2B=2 ⇒ B=1x2+1x(x2-1)=-1x+1x+1+1x-1

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The partial fractions of $\frac{x^{2} + 1}{x (x^{2} - 1)}$ are

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$- \frac{1}{x} + \frac{1}{x + 1} + \frac{1}{x - 1}$

$\frac{1}{x} - \frac{1}{x + 1} + \frac{1}{x - 1}$

$\frac{1}{x} + \frac{1}{x + 1} - \frac{1}{x - 1}$

answer is B.

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Detailed Solution

$\begin{array}{rcl} \frac{x^{2} + 1}{x (x^{2} - 1)} & = & \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x - 1} \\ x^{2} + 1 & = & A (x^{2} - 1) + Bx (x - 1) + Cx (x + 1) \\ x & = & 0 \Rightarrow A = - 1 \\ x & = & 1 \Rightarrow 2 C = 2 \Rightarrow C = 1 \\ x & = & - 1 \Rightarrow 2 B = 2 \Rightarrow B = 1 \\ \frac{x^{2} + 1}{x (x^{2} - 1)} & = & - \frac{1}{x} + \frac{1}{x + 1} + \frac{1}{x - 1} \end{array}$