Question

\( \int_0^1 \frac{1}{(x^2+16)(x^2+25)} \, dx = \)

• A. \( \frac{1}{5}\tan^{-1}\frac{1}{4} - \frac{1}{4}\tan^{-1}\frac{1}{5} \)
• B. \( \frac{1}{9}\left[\frac{1}{4}\tan^{-1}\frac{1}{4} - \frac{1}{5}\tan^{-1}\frac{1}{5}\right] \)
• C. \( \frac{1}{4}\tan^{-1}\frac{1}{4} - \frac{1}{5}\tan^{-1}\frac{1}{5} \)
• D. \( \frac{1}{9}\left[\frac{1}{5}\tan^{-1}\frac{1}{4} - \frac{1}{4}\tan^{-1}\frac{1}{5}\right] \)
• E. \( \frac{1}{9}\left[\frac{3}{4}\tan^{-1}\frac{1}{4} - \frac{4}{5}\tan^{-1}\frac{1}{5}\right] \)

Hint:

For expressions like \( (x^2+a^2)(x^2+b^2) \), always try partial fractions immediately.

Answer 1

The Correct Option is B

Understanding the Concept:
• When the integrand is a product of quadratic terms: \[ \frac{1}{(x^2+a^2)(x^2+b^2)} \] we use partial fraction decomposition.
• Then apply: \[ \int \frac{dx}{x^2+a^2} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) \]

Step 1: Set up partial fractions.
\[ \frac{1}{(x^2+16)(x^2+25)} = \frac{A}{x^2+16} + \frac{B}{x^2+25} \]

Step 2: Clear denominator.
Multiply both sides by \( (x^2+16)(x^2+25) \): \[ 1 = A(x^2+25) + B(x^2+16) \]

Step 3: Expand RHS.
\[ A(x^2+25) = Ax^2 + 25A \] \[ B(x^2+16) = Bx^2 + 16B \] Add: \[ 1 = (A+B)x^2 + (25A + 16B) \]

Step 4: Equate coefficients.
Since LHS has no \(x^2\) term: \[ A + B = 0 \quad \cdots (1) \] \[ 25A + 16B = 1 \quad \cdots (2) \]

Step 5: Solve equations.
From (1): \[ B = -A \] Substitute into (2): \[ 25A + 16(-A) = 1 \] \[ 25A - 16A = 1 \] \[ 9A = 1 \] \[ A = \frac{1}{9} \] Then: \[ B = -\frac{1}{9} \]

Step 6: Rewrite the integral.
\[ \int_0^1 \frac{1}{(x^2+16)(x^2+25)} dx \] \[ = \frac{1}{9}\int_0^1 \left( \frac{1}{x^2+16} - \frac{1}{x^2+25} \right) dx \]

Step 7: Split the integral.
\[ = \frac{1}{9} \left[ \int_0^1 \frac{dx}{x^2+16} - \int_0^1 \frac{dx}{x^2+25} \right] \]

Step 8: Apply standard formula.
\[ \int \frac{dx}{x^2+a^2} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) \] So: \[ \int \frac{dx}{x^2+16} = \frac{1}{4}\tan^{-1}\left(\frac{x}{4}\right) \] \[ \int \frac{dx}{x^2+25} = \frac{1}{5}\tan^{-1}\left(\frac{x}{5}\right) \]

Step 9: Apply limits carefully.
\[ = \frac{1}{9} \left[ \frac{1}{4}\tan^{-1}\left(\frac{x}{4}\right) - \frac{1}{5}\tan^{-1}\left(\frac{x}{5}\right) \right]_0^1 \]

Step 10: Substitute upper limit \(x=1\).
\[ = \frac{1}{9} \left[ \frac{1}{4}\tan^{-1}\frac{1}{4} - \frac{1}{5}\tan^{-1}\frac{1}{5} \right] \]

Step 11: Substitute lower limit \(x=0\).
\[ \tan^{-1}(0) = 0 \] So lower limit contributes 0.

Step 12: Final Answer.
\[ \boxed{\frac{1}{9}\left[\frac{1}{4}\tan^{-1}\frac{1}{4} - \frac{1}{5}\tan^{-1}\frac{1}{5}\right]} \]