(i) Express \(\frac{3x^2 + x}{(x+2)(x^2+1)}\) in partial fractions:

Assume \(\frac{3x^2 + x}{(x+2)(x^2+1)} = \frac{A}{x+2} + \frac{Bx+C}{x^2+1}\).

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

\(3x^2 + x = A(x^2+1) + (Bx+C)(x+2)\).

Expand and equate coefficients:

\(3x^2 + x = Ax^2 + A + Bx^2 + 2Bx + Cx + 2C\).

Combine like terms:

\((A+B)x^2 + (2B+C)x + (A+2C) = 3x^2 + x\).

Equate coefficients:

\(A + B = 3\), \(2B + C = 1\), \(A + 2C = 0\).

Solve the system of equations:

From \(A + 2C = 0\), we have \(A = -2C\).

Substitute \(A = -2C\) into \(A + B = 3\):

\(-2C + B = 3\).

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

\(2B + C = 1\).

Now solve:

From \(-2C + B = 3\), \(B = 3 + 2C\).

Substitute \(B = 3 + 2C\) into \(2B + C = 1\):

\(2(3 + 2C) + C = 1\).

\(6 + 4C + C = 1\).

\(5C = -5\).

\(C = -1\).

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

\(B = 3 + 2(-1) = 1\).

Thus, \(\frac{3x^2 + x}{(x+2)(x^2+1)} = \frac{2}{x+2} + \frac{x-1}{x^2+1}\).

(ii) Expand \(\frac{3x^2 + x}{(x+2)(x^2+1)}\) in ascending powers of \(x\):

Use the partial fractions: \(\frac{2}{x+2} + \frac{x-1}{x^2+1}\).

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

\(\frac{2}{x+2} = 2(1 + \frac{x}{2})^{-1} \approx 2(1 - \frac{x}{2} + \frac{x^2}{4} - \frac{x^3}{8})\).

Expand \(\frac{x-1}{x^2+1}\):

\(\frac{x-1}{x^2+1} = (x-1)(1 - x^2 + x^4 - \ldots) \approx (x-1)(1 - x^2)\).

\(= x - 1 - x^3 + x \approx x - 1\).

Add the expansions:

\(2(1 - \frac{x}{2} + \frac{x^2}{4} - \frac{x^3}{8}) + (x - 1)\).

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

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

Combine terms:

\(= \frac{1}{2}x + \frac{5}{4}x^2 - \frac{9}{8}x^3\).