Question

Find : \( \int \frac{x^2}{(x^2 + 9)(x^2 + 16) dx \)}

Hint:

When the numerator and denominator contain only \( x^2 \), replace \( x^2 \) with \( t \) to simplify partial fraction working, then substitute \( x^2 \) back before integrating.

Answer 1

Concept:

• Partial Fractions: Breaking down complex rational functions.
• Note: Here we only use partial fractions on \( x^2 \) algebraically, not for integration directly yet.

Step 1: Apply partial fractions algebraically
Let \( x^2 = t \). Then we consider \( \frac{t}{(t + 9)(t + 16)} \). \[ \frac{t}{(t + 9)(t + 16)} = \frac{A}{t + 9} + \frac{B}{t + 16} \] \[ t = A(t + 16) + B(t + 9) \] Putting \( t = -9 \): \( -9 = A(7) \implies A = -9/7 \). Putting \( t = -16 \): \( -16 = B(-7) \implies B = 16/7 \).

Step 2: Rewrite the integrand with \( x \)
\[ \frac{x^2}{(x^2 + 9)(x^2 + 16)} = \frac{1}{7} \left[ \frac{16}{x^2 + 16} - \frac{9}{x^2 + 9} \right] \]

Step 3: Integrate each term using standard formulas
Using \( \int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) \):
\[ I = \frac{1}{7} \left[ 16 \cdot \frac{1}{4} \tan^{-1} \left( \frac{x}{4} \right) - 9 \cdot \frac{1}{3} \tan^{-1} \left( \frac{x}{3} \right) \right] \] \[ I = \frac{1}{7} \left[ 4 \tan^{-1} \left( \frac{x}{4} \right) - 3 \tan^{-1} \left( \frac{x}{3} \right) \right] + C \]