CBSE12 Questions & Answers

Step 1: We start with the integration \( \int \frac{x^2}{(x^2 + 9)(x^2 + 16)} \, dx \).

We will use partial fraction decomposition:

$$ \frac{x^2}{(x^2 + 9)(x^2 + 16)} = \frac{Ax + B}{x^2 + 9} + \frac{Cx + D}{x^2 + 16} $$

Step 2: Multiply both sides by the denominator \((x^2 + 9)(x^2 + 16)\) to clear the fractions:

$$ x^2 = (Ax + B)(x^2 + 16) + (Cx + D)(x^2 + 9) $$

Step 3: Expand and collect like terms:

$$ x^2 = Ax^3 + 16Ax + Bx^2 + 16B + Cx^3 + 9Cx + Dx^2 + 9D $$ $$ x^2 = (A + C)x^3 + (B + D)x^2 + (16A + 9C)x + (16B + 9D) $$

Step 4: Set coefficients equal for the corresponding powers of \(x\):

• \( A + C = 0 \)
• \( B + D = 1 \)
• \( 16A + 9C = 0 \)
• \( 16B + 9D = 0 \)

Step 5: Solve the system:

• From \( A + C = 0 \) and \( 16A + 9C = 0 \), we have \( C = -A \). Substitute into the second equation:
• \( 16A + 9(-A) = 0 \) leads to \( 7A = 0 \), \(\therefore A = 0 \), \(C = 0 \)
• From \( B + D = 1 \) and \( 16B + 9D = 0 \), we can solve:
• \( D = 1 - B \), then \(16B + 9(1 - B) = 0 \) leads to \( 16B + 9 - 9B = 0 \), so \( 7B = -9 \), \(\therefore B = -\frac{9}{7} \), \(D = \frac{16}{7} \)

Step 6: Substitute back:

$$ \frac{x^2}{(x^2 + 9)(x^2 + 16)} = \frac{-\frac{9}{7}}{x^2 + 9} + \frac{\frac{16}{7}}{x^2 + 16} $$

Step 7: Integrate each term separately:

$$ \int \frac{-\frac{9}{7}}{x^2 + 9} \, dx = -\frac{9}{7} \int \frac{1}{x^2 + 9} \, dx = -\frac{9}{7} \cdot \frac{1}{3} \tan^{-1}\left(\frac{x}{3}\right) + C_1 $$ $$ \int \frac{\frac{16}{7}}{x^2 + 16} \, dx = \frac{16}{7} \int \frac{1}{x^2 + 16} \, dx = \frac{16}{7} \cdot \frac{1}{4} \tan^{-1}\left(\frac{x}{4}\right) + C_2 $$

Final Answer:

$$ \int \frac{x^2}{(x^2 + 9)(x^2 + 16)} \, dx = -\frac{3}{7} \tan^{-1}\left(\frac{x}{3}\right) + \frac{4}{7} \tan^{-1}\left(\frac{x}{4}\right) + C $$