Question

Find: \[ \int \frac{x^2}{(x^2 + 4)(x^2 + 9)} \, dx. \]

Hint:

For partial fraction decomposition, expand the equation and compare coefficients to find unknowns, then integrate each term separately.

Answer 1

Step 1: Decompose the fraction into partial fractions. \[ \frac{x^2}{(x^2 + 4)(x^2 + 9)} = \frac{A}{x^2 + 4} + \frac{B}{x^2 + 9} \] Multiplying throughout by \( (x^2 + 4)(x^2 + 9) \): \[ x^2 = A(x^2 + 9) + B(x^2 + 4) \] Expanding: \[ x^2 = (A + B)x^2 + 9A + 4B \] Equating coefficients: \[ A + B = 1, \quad 9A + 4B = 0 \tag{1} \] Step 2: Solve for \( A \) and \( B \). From equation (1): \( B = 1 - A \). Substituting in: \[ 9A + 4(1 - A) = 0 \] \[ 5A = -4 \quad \Rightarrow \quad A = -\frac{4}{5}, \quad B = \frac{9}{5} \] Step 3: Integrate each term. \[ \int \frac{x^2}{(x^2 + 4)(x^2 + 9)} \, dx = -\frac{4}{5} \int \frac{1}{x^2 + 4} \, dx + \frac{9}{5} \int \frac{1}{x^2 + 9} \, dx \] Using the standard formula: \[ \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) \] \[ = -\frac{2}{5} \tan^{-1} \left( \frac{x}{2} \right) + \frac{3}{5} \tan^{-1} \left( \frac{x}{3} \right) + C \] Final Answer: \[ \boxed{-\frac{2}{5} \tan^{-1} \left( \frac{x}{2} \right) + \frac{3}{5} \tan^{-1} \left( \frac{x}{3} \right) + C} \]