Express as partial fractions

(1)    \( \frac{x^2+x+3}{x(x^2+1)} \)

\begin{align} \frac{x^2+x+3}{x(x^2+1)} & = \frac{A}{x} + \frac{Bx+C}{(x^2+1)} \\ \\ \frac{x^2+x+3}{x(x^2+1)} & = \frac{A(x^2+1)}{x(x^2+1)} + \frac{(Bx+C)x}{(x^2+1)x} \\ \\ x^2+x+3 & = A(x^2+1) + (Bx+C)x\\ \\ & = Ax^2 + A + Bx^2 + Cx \\ \\ Equating \ constants, \quad 3 & = A \\ \\ Equating \ x^2 \ coefficients\ \quad x^2 & = Ax^2 + Bx^2 \\ 1 & = 3 + B \\ \implies \quad B & = -2 \\ \\ Equating\ x \ coefficients, \quad x & =Cx \\ \implies \quad C & = 1 \\ \\ \therefore \quad \frac{x^2+x+3}{x(x^2+1)} & = \frac{3}{x} + \frac{-2x+3}{(x^2+1)} \end{align}

(2)      \(\frac{x^2-6x+6}{x(x^2+3x-6)} \)

\begin{align} \frac{x^2-6x+6}{x(x^2+3x-6)} & = \frac{A}{x} + \frac{Bx+C}{x^2+3x-6} \\ \\ \frac{x^2-6x+6}{x(x^2+3x-6)} & = \frac{A(x^2+3x-6)}{x(x^2+3x-6)} + \frac{(Bx+C)x}{x(x^2+3x-6)} \\ \\ x^2-6x+6 & = A(x^2+3x-6) + (Bx+C)x \\ \\ & = (A+B)x^2+(3A+C)x -6A \\ \\ Equating \ constants, \quad 6 & = -6A \\ \implies \quad A & =-1 \\ \\ Equating \ x^2 \ coefficients, \quad x^2 & = (-1+B)x^2 \\ \implies \quad B & =2 \\ \\ Equating \ x \ coefficients, \quad -6x & = (-3+C)x \\ \implies \quad C & =-3 \\ \\ \therefore \quad \frac{x^2-6x+6}{x(x^2+3x-6)} & = \frac{-1}{x} + \frac{2x-3}{x^2+3x-6} \\ \end{align}

(3)    \( \frac{3x^3+7x-4}{(x^2+2)^2} \)

\begin{align} \frac{3x^3+7x-4}{(x^2+2)^2} & = \frac{Ax+B}{x^2+2} + \frac{Cx+D}{(x^2+2)^2} \\ \\ 3x^3+7x-4 & = (Ax+B)(x^2+2) + (Cx+D) \\ \\ 3x^3+7x-4 & = Ax^3+Bx^2 +2Ax +2B + Cx+D\\ \\ & = Ax^3 + Bx^2 + (2A+C)x + 2B+D \\ \\ Equating \ x^3 \ coefficients\ \quad 3x^3 & = Ax^3 \\ \implies \quad A & = 3 \\ \\ Equating \ x^2 \ coefficients\ \quad 0x^2 & = Bx^2 \\ \implies \quad B & = 0 \\ \\ Equating \ x \ coefficients, \quad 7x & = (2A+C)x \\ 7 & = 2(3)+C \\ \implies \quad C=1 \\ Equating\ constants, \quad -4 & = 2B+D \\ \implies \quad D & = -4 \\ \\ \therefore \quad \frac{3x^3+7x-4}{(x^2+2)^2} & = \frac{3x}{x^2+2} + \frac{x-4}{(x^2+2)^2} \end{align}