Question

If $\int \frac{2x^{2}+a^{2}}{x^{2}(x^{2}+a^{2})} dx = \frac{k}{x} + \frac{1}{a} \tan^{-1} \frac{x}{a} + c$ then $k =$

• A. 0
• B. -1
• C. 1
• D. $1/a$

Hint:

Splitting fractions is often much faster than using the partial fractions formula.

Answer 1

The Correct Option is B

Step 1: Concept
Split the numerator into parts that match the denominator terms.

Step 2: Meaning
$\frac{2x^{2}+a^{2}}{x^{2}(x^{2}+a^{2})} = \frac{(x^{2}+a^{2}) + x^{2}}{x^{2}(x^{2}+a^{2})} = \frac{1}{x^{2}} + \frac{1}{x^{2}+a^{2}}$.

Step 3: Analysis
Integrating each term: $\int \frac{1}{x^{2}} dx + \int \frac{1}{x^{2}+a^{2}} dx = -\frac{1}{x} + \frac{1}{a} \tan^{-1} \frac{x}{a}$.

Step 4: Conclusion
Comparing this with $\frac{k}{x} + \frac{1}{a} \tan^{-1} \frac{x}{a}$, we find $k = -1$. Final Answer: (B)