Question

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

Hint:

To solve rational integrals, use partial fraction decomposition to split the integrand into simpler terms, then integrate using standard formulas.

Answer 1

To solve the given integral, we use partial fraction decomposition. Let: \[ \frac{x^2}{(x^2 + 4)(x^2 + 9)} = \frac{A}{x^2 + 4} + \frac{B}{x^2 + 9}. \] Multiply through by \((x^2 + 4)(x^2 + 9)\) to eliminate the denominators: \[ x^2 = A(x^2 + 9) + B(x^2 + 4). \] Simplify: \[ x^2 = A x^2 + 9A + B x^2 + 4B. \] Combine like terms: \[ x^2 = (A + B)x^2 + (9A + 4B). \] Equating coefficients of \(x^2\) and the constant term: \[ A + B = 1, \quad 9A + 4B = 0. \tag{1} \] From the first equation: \[ B = 1 - A. \tag{2} \] Substitute \(B = 1 - A\) into the second equation: \[ 9A + 4(1 - A) = 0. \] Simplify: \[ 9A + 4 - 4A = 0, \] \[ 5A = -4, \quad A = -\frac{4}{5}. \] Substitute \(A = -\frac{4}{5}\) into \(B = 1 - A\): \[ B = 1 - \left(-\frac{4}{5}\right) = 1 + \frac{4}{5} = \frac{9}{5}. \] Thus: \[ \frac{x^2}{(x^2 + 4)(x^2 + 9)} = \frac{-\frac{4}{5}}{x^2 + 4} + \frac{\frac{9}{5}}{x^2 + 9}. \] Rewrite: \[ \frac{x^2}{(x^2 + 4)(x^2 + 9)} = -\frac{4}{5} \cdot \frac{1}{x^2 + 4} + \frac{9}{5} \cdot \frac{1}{x^2 + 9}. \] The integral becomes: \[ \int \frac{x^2}{(x^2 + 4)(x^2 + 9)} \, dx = -\frac{4}{5} \int \frac{1}{x^2 + 4} \, dx + \frac{9}{5} \int \frac{1}{x^2 + 9} \, dx. \] Using the standard formula: \[ \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right), \] we evaluate each term: 1. For \(\int \frac{1}{x^2 + 4} \, dx\): \[ \int \frac{1}{x^2 + 4} \, dx = \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right). \] 2. For \(\int \frac{1}{x^2 + 9} \, dx\): \[ \int \frac{1}{x^2 + 9} \, dx = \frac{1}{3} \tan^{-1}\left(\frac{x}{3}\right). \] Substitute these results back: \[ \int \frac{x^2}{(x^2 + 4)(x^2 + 9)} \, dx = -\frac{4}{5} \cdot \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + \frac{9}{5} \cdot \frac{1}{3} \tan^{-1}\left(\frac{x}{3}\right). \] Simplify: \[ \int \frac{x^2}{(x^2 + 4)(x^2 + 9)} \, dx = -\frac{2}{5} \tan^{-1}\left(\frac{x}{2}\right) + \frac{3}{5} \tan^{-1}\left(\frac{x}{3}\right) + C, \] where \(C\) is the constant of integration. Final Answer: \[ \int \frac{x^2}{(x^2 + 4)(x^2 + 9)} \, dx = -\frac{2}{5} \tan^{-1}\left(\frac{x}{2}\right) + \frac{3}{5} \tan^{-1}\left(\frac{x}{3}\right) + C. \]