Question

\(∫\frac{dx}{(x2+1) (x2+4)} =\)

• A. \(\frac{1}{3}Tan^{-1}x +\frac{1}{6}Tan^{-1}(\frac{x}{2})+c\)
• B. \(\frac{1}{3}Tan^{-1}x -\frac{1}{3}Tan^{-1}(\frac{x}{2})+c\)
• C. \(\frac{1}{3}Tan^{-1}x +\frac{1}{3}Tan^{-1}(\frac{x}{2})+c\)
• D. \(\frac{1}{3}Tan^{-1}x -\frac{1}{6}Tan^{-1}(\frac{x}{2})+c\)

Answer 1

The Correct Option is D

To solve the problem, we need to evaluate the integral $I = \int \frac{dx}{(x^2 + 1)(x^2 + 4)}$.

1. Decompose the Integrand:
Write $\frac{1}{(x^2 + 1)(x^2 + 4)} = \frac{A}{x^2 + 1} + \frac{B}{x^2 + 4}$.
Multiply through by $(x^2 + 1)(x^2 + 4)$:
$1 = A(x^2 + 4) + B(x^2 + 1) = (A + B)x^2 + (4A + B)$.
Equate coefficients:
$A + B = 0$, $4A + B = 1$.

2. Solve for Coefficients:
Subtract $A + B = 0$ from $4A + B = 1$:
$3A = 1$, so $A = \frac{1}{3}$.
Then $B = -A = -\frac{1}{3}$.

3. Rewrite the Integral:
Substitute back:
$I = \int \left( \frac{\frac{1}{3}}{x^2 + 1} - \frac{\frac{1}{3}}{x^2 + 4} \right) dx = \frac{1}{3} \int \frac{dx}{x^2 + 1} - \frac{1}{3} \int \frac{dx}{x^2 + 4}$.

4. Evaluate the Integrals:
The first integral is:
$\int \frac{dx}{x^2 + 1} = \arctan x$.
The second integral is:
$\int \frac{dx}{x^2 + 4} = \int \frac{dx}{x^2 + 2^2} = \frac{1}{2} \arctan \frac{x}{2}$.
Thus:
$I = \frac{1}{3} \arctan x - \frac{1}{3} \cdot \frac{1}{2} \arctan \frac{x}{2} + C = \frac{1}{3} \arctan x - \frac{1}{6} \arctan \frac{x}{2} + C$.

Final Answer:
The integral is $\frac{1}{3} \arctan x - \frac{1}{6} \arctan \frac{x}{2} + C$.