Question

Evaluate the integral \( \displaystyle \int \frac{x^2}{(x^2 + 1)(x^2 + 4)} \, dx \) using partial fractions.

Hint:

For quadratic denominators: \( \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}(x/a) \). Always split using partial fractions first.

Answer 1

Concept: To integrate rational functions, express the integrand using partial fractions and then use standard inverse trigonometric integrals: \[ \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\!\left(\frac{x}{a}\right) \] Step 1: Use partial fractions.
Let: \[ \frac{x^2}{(x^2 + 1)(x^2 + 4)} = \frac{A}{x^2 + 1} + \frac{B}{x^2 + 4} \]
Step 2: Clear denominators.
\[ x^2 = A(x^2 + 4) + B(x^2 + 1) \] \[ x^2 = (A + B)x^2 + (4A + B) \]
Step 3: Compare coefficients.
\[ A + B = 1 \quad \cdots (1) \] \[ 4A + B = 0 \quad \cdots (2) \]
Step 4: Solve equations.
From (1): \( B = 1 - A \) Substitute into (2): \[ 4A + (1 - A) = 0 \] \[ 3A = -1 \Rightarrow A = -\frac{1}{3} \] \[ B = 1 + \frac{1}{3} = \frac{4}{3} \]

Step 5: Rewrite the integral.
\[ \int \left( \frac{-1/3}{x^2 + 1} + \frac{4/3}{x^2 + 4} \right) dx \] \[ = -\frac{1}{3} \int \frac{dx}{x^2 + 1} + \frac{4}{3} \int \frac{dx}{x^2 + 4} \]
Step 6: Use standard formulas.
\[ \int \frac{dx}{x^2 + 1} = \tan^{-1}x \] \[ \int \frac{dx}{x^2 + 4} = \frac{1}{2}\tan^{-1}\!\left(\frac{x}{2}\right) \]
Step 7: Substitute results.
\[ = -\frac{1}{3}\tan^{-1}x + \frac{4}{3} \cdot \frac{1}{2}\tan^{-1}\!\left(\frac{x}{2}\right) \] \[ = -\frac{1}{3}\tan^{-1}x + \frac{2}{3}\tan^{-1}\!\left(\frac{x}{2}\right) \] Rearranging: \[ = \frac{1}{3}\tan^{-1}x - \frac{1}{6}\tan^{-1}\!\left(\frac{x}{2}\right) + C \] Conclusion:
\[ \int \frac{x^2}{(x^2 + 1)(x^2 + 4)} dx = \frac{1}{3}\tan^{-1}x - \frac{1}{6}\tan^{-1}\!\left(\frac{x}{2}\right) + C \]