To evaluate the integral: \[ \int \frac{2x}{(x^2 + 3)(x^2 - 5)} \, dx \] We apply partial fraction decomposition. The integrand is expressed as: \[ \frac{2x}{(x^2 + 3)(x^2 - 5)} = \frac{A}{x^2 + 3} + \frac{B}{x^2 - 5} \] Multiplying by \( (x^2 + 3)(x^2 - 5) \) yields: \[ 2x = A(x^2 - 5) + B(x^2 + 3) \] Expanding this equation: \[ 2x = A x^2 - 5A + B x^2 + 3B \] Grouping terms by powers of \( x \): \[ 2x = (A + B) x^2 + (-5A + 3B) \] Equating coefficients of like powers of \( x \) on both sides. For \( x^2 \): \[ A + B = 0 \] For \( x \): \[ -5A + 3B = 2 \] Solving the system of equations: From \( A + B = 0 \), we get \( B = -A \). Substituting this into the second equation: \[ -5A + 3(-A) = 2 \] \[ -5A - 3A = 2 \] \[ -8A = 2 \] \[ A = -\frac{1}{4} \] Since \( B = -A \), we find: \[ B = \frac{1}{4} \] The partial fraction decomposition is therefore: \[ \frac{2x}{(x^2 + 3)(x^2 - 5)} = \frac{-1/4}{x^2 + 3} + \frac{1/4}{x^2 - 5} \] Integrating each term: \[ \int \frac{-1/4}{x^2 + 3} \, dx = -\frac{1}{4} \int \frac{1}{x^2 + 3} \, dx = -\frac{1}{4} \cdot \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{x}{\sqrt{3}} \right) \] \[ \int \frac{1/4}{x^2 - 5} \, dx = \frac{1}{4} \int \frac{1}{x^2 - 5} \, dx = \frac{1}{4} \cdot \frac{1}{\sqrt{5}} \tanh^{-1} \left( \frac{x}{\sqrt{5}} \right) \] The final result of the integration is: \[ \int \frac{2x}{(x^2 + 3)(x^2 - 5)} \, dx = -\frac{1}{4\sqrt{3}} \tan^{-1} \left( \frac{x}{\sqrt{3}} \right) + \frac{1}{4\sqrt{5}} \tanh^{-1} \left( \frac{x}{\sqrt{5}} \right) + C \]