Question

Find: \[ \int \frac{x^2 + 1}{(x^2 + 2)(x^2 + 4)} \, dx \]

Hint:

For rational functions, use partial fraction decomposition to simplify integration.

Answer 1

1. Decompose into partial fractions: Let: \[ \frac{x^2 + 1}{(x^2 + 2)(x^2 + 4)} = \frac{A}{x^2 + 2} + \frac{B}{x^2 + 4}. \] Multiply through by \( (x^2 + 2)(x^2 + 4) \): \[ x^2 + 1 = A(x^2 + 4) + B(x^2 + 2). \] 2. Expand and compare coefficients: \[ x^2 + 1 = A x^2 + 4A + B x^2 + 2B = (A + B)x^2 + (4A + 2B). \] Equating coefficients of \( x^2 \) and constant terms: \[ A + B = 1, \quad 4A + 2B = 1. \] Solve for \( A \) and \( B \): \[ A + B = 1 \quad \Rightarrow \quad B = 1 - A, \] \[ 4A + 2(1 - A) = 1 \quad \Rightarrow \quad 4A + 2 - 2A = 1 \quad \Rightarrow \quad 2A = -1 \quad \Rightarrow \quad A = -\frac{1}{2}. \] Substitute \( A \) into \( A + B = 1 \): \[ -\frac{1}{2} + B = 1 \quad \Rightarrow \quad B = \frac{3}{2}. \] 3. Rewrite the integral: \[ \int \frac{x^2 + 1}{(x^2 + 2)(x^2 + 4)} \, dx = \int \frac{-\frac{1}{2}}{x^2 + 2} \, dx + \int \frac{\frac{3}{2}}{x^2 + 4} \, dx. \] 4. Integrate: \[ \int \frac{-\frac{1}{2}}{x^2 + 2} \, dx = -\frac{1}{2} \cdot \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x}{\sqrt{2}}\right), \] \[ \int \frac{\frac{3}{2}}{x^2 + 4} \, dx = \frac{3}{2} \cdot \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right). \] Combine: \[ \int \frac{x^2 + 1}{(x^2 + 2)(x^2 + 4)} \, dx = -\frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{x}{\sqrt{2}}\right) + \frac{3}{4} \tan^{-1}\left(\frac{x}{2}\right) + C. \] Final Answer: \[ -\frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{x}{\sqrt{2}}\right) + \frac{3}{4} \tan^{-1}\left(\frac{x}{2}\right) + C \]