\begin{align*} {x^2 + 2x + 15 \over x(x^2 + 3)} & = {A \over x} + {Bx + C \over x^2 + 3} \phantom{000000000000000} [\text{Step 1}] \\ & = {A(x^2 + 3) \over x(x^2 + 3)} + {(Bx + C)(x) \over x(x^2 + 3)} \\ & = {A(x^2 + 3) + (Bx + C)(x) \over x(x^2 + 3)} \\ \\ \text{Comparing } & \text{the numerator,} \\ x^2 + 2x + 15 & = A(x^2 + 3) + (Bx + C)(x) \phantom{000 000} [\text{Step 2}] \\ \\ \text{Let } & x = 0, \\ (0)^2 + 2(0) + 15 & = A[(0)^2 + 3] + [B(0) + C](0) \\ 15 & = A(3) + (C)(0) \\ 15 & = 3A \\ {15 \over 3} & = A \\ 5 & = A \\ \\ x^2 + 2x + 15 & = 5(x^2 + 3) + (Bx + C)(x) \\ x^2 + 2x + 15 & = 5x^2 + 15 + Bx^2 + Cx \\ x^2 + 2x + 15 & = 5x^2 + Bx^2 + Cx + 15 \\ x^2 + 2x + 15 & = (5 + B)x^2 + Cx + 15 \\ \\ \text{Comparing } & \text{coefficients of } x^2, \\ 1 & = 5 + B \\ 1 - 5 & = B \\ -4 & = B \\ \\ \text{Comparing } & \text{coefficients of } x, \\ 2 & = C \\ \\ \therefore {x^2 + 2x + 15 \over x(x^2 + 3)} & = {A \over x} + {Bx + C \over x^2 + 3} \\ & = {5 \over x} + {(-4)x + (2) \over x^2 + 3} \\ & = {5 \over x} + {2 - 4x \over x^2 + 3} \end{align*}

\begin{align} \text{Let } g(x) & = 2x^3 + 5x^2 + 4x + 1 \\ \\ g(-1) & = 2(-1)^3 + 5(-1)^2 + 4(-1) + 1 \\ & = 0 \\ \\ \therefore x + 1 & \text{ is a factor of } g(x) \end{align}
$$ \require{enclose} \begin{array}{rll} 2x^2 + 3x + 1 \phantom{000000}\\ x + 1 \enclose{longdiv}{ 2x^3 + 5x^2 + 4x + 1 \phantom{0}}\kern-.2ex \\ -\underline{( 2x^3 + 2x^2){\phantom{00000000}}} \\ 3x^2 + 4x + 1 \phantom{0} \\ -\underline{( 3x^2 + 3x){\phantom{000.}}} \\ x + 1 \phantom{0} \\ -\underline{(x + 1){\phantom{.}}} \\ 0 \phantom{0} \end{array} $$
\begin{align} 2x^3 + 5x^2 + 4x + 1 & = (x + 1)(2x^2 + 3x + 1) \\ & = (x + 1)(x + 1)(2x + 1) \\ & = (x + 1)^2 (2x + 1) \end{align}

\begin{align} {6 + 10x - x^2 \over 2x^3 + 5x + 4x + 1} & = {6 + 10x - x^2 \over (2x + 1)(x + 1)^2} \phantom{000000} [\text{Use factorisation from (i)}] \\ & = {A \over 2x + 1} + {B \over x + 1} + {C \over (x + 1)^2} \\ & = {A(x + 1)^2 \over (2x + 1)(x + 1)^2} + {B(2x + 1)(x + 1) \over (2x + 1)(x + 1)^2} + {C(2x + 1) \over (2x + 1)(x + 1)^2} \\ & = {A(x + 1)^2 + B(2x + 1)(x + 1) + C(2x + 1) \over (2x + 1)(x + 1)^2} \\ \\ 6 + 10x - x^2 & = A(x + 1)^2 + B(2x + 1)(x + 1) + C(2x + 1) \\ \\ \text{Let } & x = -1, \\ 6 + 10(-1) - (-1)^2 & = 0 + 0 + C[2(-1) + 1] \\ -5 & = -C \\ 5 & = C \\ \\ 6 + 10x - x^2 & = A(x + 1)^2 + B(2x + 1)(x + 1) + 5(2x + 1) \\ \\ \text{Let } & x = -0.5, \\ 6 + 10(-0.5) - (-0.5)^2 & = A(0.5)^2 + 0 + 0 \\ 0.75 & = 0.25A \\ {0.75 \over 0.25} & = A \\ 3 & = A \\ \\ 6 + 10x - x^2 & = 3(x + 1)^2 + B(2x + 1)(x + 1) + 5(2x + 1) \\ \\ \text{Let } & x = 0, \\ 6 + 0 - 0 & = 3(1)^2 + B(1)(1) + 5(1) \\ 6 & = 3 + B + 5 \\ -B & = 3 + 5 - 6 \\ -B & = 2 \\ B & = -2 \\ \\ \\ \therefore {6 + 10x - x^2 \over (2x + 1)(x + 1)^2} & = {3 \over 2x + 1} + {-2 \over x + 1} + {5 \over (x + 1)^2} \\ & = {3 \over 2x + 1} - {2 \over x + 1} + {5 \over (x + 1)^2} \end{align}

$$ \require{enclose} \begin{array}{rll} x \phantom{0000000000000}\\ x^2 + 0x - 1 \enclose{longdiv}{ x^3 + 0x^2 - x + 4 \phantom{0}}\kern-.2ex \\ -\underline{( x^3 + 0x^2 - x){\phantom{000.}}} \\ 4 \phantom{0} \end{array} $$
\begin{align} {x^3 - x + 4 \over x^2 - 1} & = x + \underbrace{ {4 \over x^2 - 1} }_\text{Proper fraction} \\ \\ {4 \over x^2 - 1} & = {4 \over (x + 1)(x - 1)} \phantom{0000000} [a^2 - b^2 = (a + b)(a - b)] \\ & = {A \over x + 1} + {B \over x - 1} \\ & = {A(x - 1) \over (x + 1)(x - 1)} + {B(x + 1) \over (x + 1)(x - 1)} \\ & = {A(x - 1) + B(x + 1) \over (x + 1)(x - 1)} \\ \\ 4 & = A(x - 1) + B(x + 1) \\ \\ \text{Let } & x = 1, \\ 4 & = 0 + B(2) \\ 4 & = 2B \\ {4 \over 2} & = B \\ 2 & = B \\ \\ \text{Let } & x = -1, \\ 4 & = A(-2) + 0 \\ 4 & = -2A \\ {4 \over -2} & = A \\ -2 & = A \\ \\ {4 \over x^2 - 1} & = {-2 \over x + 1} + {2 \over x - 1} \\ & = -{2 \over x + 1} + {2 \over x - 1} \\ \\ \\ \therefore {x^3 - x + 4 \over x^2 - 1} & = x - {2 \over x + 1} + {2 \over x - 1} \end{align}