(i) Express \(\frac{5x - x^2}{(1+x)(2+x^2)}\) in partial fractions:

Assume \(\frac{5x - x^2}{(1+x)(2+x^2)} = \frac{A}{1+x} + \frac{Bx + C}{2+x^2}\).

Multiply through by \((1+x)(2+x^2)\) to clear the denominators:

\(5x - x^2 = A(2+x^2) + (Bx + C)(1+x)\).

Expand and collect like terms:

\(5x - x^2 = 2A + Ax^2 + Bx + Bx^2 + C + Cx\).

Equate coefficients:

For \(x^2\): \(-1 = A + B\)

For \(x\): \(5 = B + C\)

For constant term: \(0 = 2A + C\)

Solve the system of equations:

1. \(A + B = -1\)

2. \(B + C = 5\)

3. \(2A + C = 0\)

From equation 3: \(C = -2A\).

Substitute \(C = -2A\) into equation 2: \(B - 2A = 5\).

Substitute \(B = -1 - A\) from equation 1 into \(B - 2A = 5\):

\(-1 - A - 2A = 5\)

\(-1 - 3A = 5\)

\(-3A = 6\)

\(A = -2\)

Substitute \(A = -2\) into \(B = -1 - A\):

\(B = -1 + 2 = 1\)

Substitute \(A = -2\) into \(C = -2A\):

\(C = 4\)

Thus, \(\frac{5x - x^2}{(1+x)(2+x^2)} = \frac{-2}{1+x} + \frac{x + 4}{2+x^2}\).

(ii) Expand \(\frac{5x - x^2}{(1+x)(2+x^2)}\) in ascending powers of \(x\):

Expand \(\frac{-2}{1+x}\) using the binomial series:

\(-2(1 - x + x^2 - x^3 + \cdots) = -2 + 2x - 2x^2 + 2x^3 + \cdots\)

Expand \(\frac{x + 4}{2+x^2}\) using the binomial series:

\((x + 4)(\frac{1}{2} - \frac{x^2}{4} + \cdots) = \frac{x}{2} + 2 - \frac{x^3}{4} + \cdots\)

Combine the expansions:

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

Combine like terms:

\(\frac{5}{2}x - 3x^2 + \frac{7}{4}x^3 + \cdots\)