Question

If $\int \frac{dx}{x^4 + 5x^2 + 4} = A \tan^{-1} x + B \tan^{-1} \frac{x}{2} + c$ where $c$ is a constant of integration, then ______.

• A. $A = \frac{1}{2}$, $B = \frac{1}{4}$
• B. $A = \frac{1}{3}$, $B = -\frac{1}{6}$
• C. $A = \frac{1}{3}$, $B = \frac{1}{6}$
• D. $A = \frac{1}{2}$, $B = -\frac{1}{4}$

Hint:

To quickly split $\frac{1}{(x^2+a)(x^2+b)}$, use the constant difference trick: $\frac{1}{(x^2+a)(x^2+b)} = \frac{1}{b-a} \left[ \frac{1}{x^2+a} - \frac{1}{x^2+b} \right]$. Here, $b=4, a=1$, so coefficient is $1/(4-1) = 1/3$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The integrand is a rational function. We can use partial fractions by treating $x^2$ as a single variable (say $t$) to decompose the fraction.

Step 2: Formula Application:
$\frac{1}{(x^2+1)(x^2+4)} = \frac{1}{3} \left( \frac{1}{x^2+1} - \frac{1}{x^2+4} \right)$.

Step 3: Explanation:
$\int \frac{dx}{(x^2+1)(x^2+4)} = \frac{1}{3} \int \frac{1}{x^2+1} \, dx - \frac{1}{3} \int \frac{1}{x^2+4} \, dx$. Integrating both terms: $\frac{1}{3} \tan^{-1} x - \frac{1}{3} \left( \frac{1}{2} \tan^{-1} \frac{x}{2} \right) + c$. This simplifies to $\frac{1}{3} \tan^{-1} x - \frac{1}{6} \tan^{-1} \frac{x}{2} + c$. Comparing with the given form, $A = \frac{1}{3}$ and $B = -\frac{1}{6}$.

Step 4: Final Answer:
$A = \frac{1}{3}$ and $B = -\frac{1}{6}$.