Question

\( \int \frac{x^2 + 4x^4 + 16}{dx} \) is equal to

• A. $\frac{1}{2\sqrt{2}}tan^{-1}(\frac{x^{2}+4}{2x})+c$
• B. $\frac{1}{2\sqrt{2}}tan^{-1}(\frac{x^{2}-4}{2\sqrt{2}x})+c$
• C. $\frac{1}{2\sqrt{2}}tan^{-1}(\frac{x^{2}+4}{2\sqrt{2}x})+c$
• D. $\frac{1}{2}tan^{-1}(\frac{x^{2}-4}{2x})+c$

Hint:

$\intx dx$ is equal to

Answer 1

The Correct Option is B

Step 1: Concept
Divide numerator and denominator by $x^2$.
Step 2: Analysis
The integral becomes $\int \frac{1 + 4/x^2}{x^2 + 16/x^2} \, dx$.
Step 3: Evaluation
Rewrite the denominator: $x^2 + 16/x^2 = (x - 4/x)^2 + 8$. Let $t = x - 4/x$, then $dt = (1 + 4/x^2) \, dx$.
Step 4: Conclusion
The integral becomes $\int \frac{dt}{t^2 + (2\sqrt{2})^2} = \frac{1}{2\sqrt{2}} tan^{-1}(\frac{t}{2\sqrt{2}}) + c$. Substituting $t$ back gives the result.
Final Answer: (b)