To evaluate the given integral, we employ partial fraction decomposition.We set up the decomposition as follows:\[\frac{x^2}{(x^2 + 4)(x^2 + 9)} = \frac{A}{x^2 + 4} + \frac{B}{x^2 + 9}.\]Multiplying both sides by \((x^2 + 4)(x^2 + 9)\) to clear denominators yields:\[x^2 = A(x^2 + 9) + B(x^2 + 4).\]Expanding and simplifying the right side:\[x^2 = A x^2 + 9A + B x^2 + 4B.\]Grouping like terms:\[x^2 = (A + B)x^2 + (9A + 4B).\]Equating the coefficients of \(x^2\) and the constant terms, we obtain a system of equations:\[A + B = 1, \quad 9A + 4B = 0. \tag{1}\]From the first equation, we express \(B\) in terms of \(A\):\[B = 1 - A. \tag{2}\]Substituting this expression for \(B\) into the second equation:\[9A + 4(1 - A) = 0.\]Simplifying this equation:\[9A + 4 - 4A = 0,\]which leads to:\[5A = -4, \quad A = -\frac{4}{5}.\]Substituting the value of \(A\) back into the equation for \(B\):\[B = 1 - \left(-\frac{4}{5}\right) = 1 + \frac{4}{5} = \frac{9}{5}.\]Thus, the partial fraction decomposition is:\[\frac{x^2}{(x^2 + 4)(x^2 + 9)} = \frac{-\frac{4}{5}}{x^2 + 4} + \frac{\frac{9}{5}}{x^2 + 9}.\]This can be rewritten as:\[\frac{x^2}{(x^2 + 4)(x^2 + 9)} = -\frac{4}{5} \cdot \frac{1}{x^2 + 4} + \frac{9}{5} \cdot \frac{1}{x^2 + 9}.\]The integral transforms into:\[\int \frac{x^2}{(x^2 + 4)(x^2 + 9)} \, dx = -\frac{4}{5} \int \frac{1}{x^2 + 4} \, dx + \frac{9}{5} \int \frac{1}{x^2 + 9} \, dx.\]Using the standard integral formula \(\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)\), we evaluate each integral separately:1. For the term \(\int \frac{1}{x^2 + 4} \, dx\):\[\int \frac{1}{x^2 + 4} \, dx = \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right).\]2. For the term \(\int \frac{1}{x^2 + 9} \, dx\):\[\int \frac{1}{x^2 + 9} \, dx = \frac{1}{3} \tan^{-1}\left(\frac{x}{3}\right).\]Substituting these results back into the main integral:\[\int \frac{x^2}{(x^2 + 4)(x^2 + 9)} \, dx = -\frac{4}{5} \cdot \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + \frac{9}{5} \cdot \frac{1}{3} \tan^{-1}\left(\frac{x}{3}\right).\]Simplifying the expression:\[\int \frac{x^2}{(x^2 + 4)(x^2 + 9)} \, dx = -\frac{2}{5} \tan^{-1}\left(\frac{x}{2}\right) + \frac{3}{5} \tan^{-1}\left(\frac{x}{3}\right) + C,\]where \(C\) denotes the constant of integration. Final Answer:\[\int \frac{x^2}{(x^2 + 4)(x^2 + 9)} \, dx = -\frac{2}{5} \tan^{-1}\left(\frac{x}{2}\right) + \frac{3}{5} \tan^{-1}\left(\frac{x}{3}\right) + C.\]