Problem Statement
Evaluate the indefinite integral:
$$\int \frac{x^2 + 1}{(x^2 + 4)(x^2 + 9)} \, dx$$
Verified Solution & Marking Scheme
Temporary algebraic substitution for partial fraction decomposition
Put $x^2 = y$ only to determine partial fractions:
$$\frac{y + 1}{(y + 4)(y + 9)} = \frac{A}{y + 4} + \frac{B}{y + 9}$$
$$y + 1 = A(y + 9) + B(y + 4)$$
Determine coefficients A and B
Put $y = -4$: $-4 + 1 = A(5) \implies -3 = 5A \implies A = -\frac{3}{5}$.
Put $y = -9$: $-9 + 1 = B(-5) \implies -8 = -5B \implies B = \frac{8}{5}$.
Restore x² and integrate using standard formula ∫ dx/(x² + a²)
$$\frac{x^2 + 1}{(x^2 + 4)(x^2 + 9)} = -\frac{3}{5} \cdot \frac{1}{x^2 + 4} + \frac{8}{5} \cdot \frac{1}{x^2 + 9}$$
Integrating with respect to $x$:
$$\int \frac{x^2 + 1}{(x^2 + 4)(x^2 + 9)} \, dx = -\frac{3}{5} \int \frac{dx}{x^2 + 2^2} + \frac{8}{5} \int \frac{dx}{x^2 + 3^2}$$
$$= -\frac{3}{5} \cdot \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + \frac{8}{5} \cdot \frac{1}{3} \tan^{-1}\left(\frac{x}{3}\right) + C$$
$$= -\frac{3}{10} \tan^{-1}\left(\frac{x}{2}\right) + \frac{8}{15} \tan^{-1}\left(\frac{x}{3}\right) + C$$