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 Question:
We must integrate a rational function by factoring the denominator and applying partial fractions, then compare the result to the given expression to identify constants A and B.

Step 2: Detailed Explanation:
Let the integral be $I = \int \frac{1}{x^4 + 5x^2 + 4} dx$.
Treat $x^2$ temporarily as a variable $t$ to factor the denominator:
$t^2 + 5t + 4 = (t + 1)(t + 4)$
Thus, $x^4 + 5x^2 + 4 = (x^2 + 1)(x^2 + 4)$.
Now, use partial fraction decomposition on the integrand:
$\frac{1}{(x^2 + 1)(x^2 + 4)} = \frac{P}{x^2 + 1} + \frac{Q}{x^2 + 4}$
Using the cover-up method or algebraic solving:
$1 = P(x^2 + 4) + Q(x^2 + 1)$
If $x^2 = -1 \implies 1 = P(-1 + 4) \implies 3P = 1 \implies P = \frac{1}{3}$
If $x^2 = -4 \implies 1 = Q(-4 + 1) \implies -3Q = 1 \implies Q = -\frac{1}{3}$
Substitute $P$ and $Q$ back into the integral:
$I = \int \left( \frac{1/3}{x^2 + 1} - \frac{1/3}{x^2 + 4} \right) dx$
$I = \frac{1}{3} \int \frac{1}{x^2 + 1^2} dx - \frac{1}{3} \int \frac{1}{x^2 + 2^2} dx$
Use the standard integration formula $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)$:
For the first term: $a = 1 \implies \frac{1}{3} \cdot \frac{1}{1} \tan^{-1}(x) = \frac{1}{3} \tan^{-1}(x)$
For the second term: $a = 2 \implies -\frac{1}{3} \cdot \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) = -\frac{1}{6} \tan^{-1}\left(\frac{x}{2}\right)$
Combining them:
$I = \frac{1}{3} \tan^{-1}(x) - \frac{1}{6} \tan^{-1}\left(\frac{x}{2}\right) + c$
Comparing this to the given expression $A \tan^{-1} x + B \tan^{-1} \frac{x}{2} + c$:
$A = \frac{1}{3}$
$B = -\frac{1}{6}$

Step 3: Final Answer:
The values are $A = \frac{1}{3}, B = -\frac{1}{6}$, matching option (b).