(i) To express \(\frac{100}{x^2(10-x)}\) in partial fractions, assume the form:
\(\frac{A}{x} + \frac{B}{x^2} + \frac{C}{10-x}\)
Multiply through by the denominator \(x^2(10-x)\) to get:
\(100 = A x (10-x) + B (10-x) + C x^2\)
Equating coefficients, solve for \(A, B,\) and \(C\):
\(A = 1, B = 10, C = 1\)
Thus, \(\frac{100}{x^2(10-x)} = \frac{1}{x} + \frac{10}{x^2} + \frac{1}{10-x}\).
(ii) Separate variables and integrate:
\(\int \frac{1}{x^2(10-x)} \, dx = \int \frac{1}{100} \, dt\)
Integrate the left side using partial fractions:
\(\int \left( \frac{1}{x} + \frac{10}{x^2} + \frac{1}{10-x} \right) \, dx\)
\(= \ln |x| - \frac{10}{x} - \ln |10-x|\)
Integrate the right side:
\(\frac{t}{100} + C\)
Apply initial conditions \(x = 1, t = 0\):
\(\ln 1 - \frac{10}{1} - \ln 9 = 0 + C\)
\(C = -10 - \ln 9\)
Substitute back to find \(t\):
\(\ln |x| - \frac{10}{x} - \ln |10-x| = \frac{t}{100} - 10 - \ln 9\)
Simplify to obtain:
\(t = \ln \left( \frac{9x}{10-x} \right) - \frac{10}{x} + 10\)
