Step 1: Understanding the Concept:
This integral belongs to a specific class of rational functions where the denominator is $x^4 + kx^2 + 1$.
The technique used is to reduce the degree of the terms by dividing both the numerator and the denominator by $x^2$.
Once divided, we aim to transform the denominator into a perfect square plus or minus a constant, which aligns with the derivative of a chosen substitution.
If the numerator becomes $1 + \frac{1}{x^2}$, we use the substitution $u = x - \frac{1}{x}$ because its derivative is exactly the numerator.
Step 2: Key Formula or Approach:
1. Divide numerator and denominator by $x^2$.
2. Rewrite the denominator $x^2 + \frac{1}{x^2}$ as $(x - \frac{1}{x})^2 + 2$.
3. Let $u = x - \frac{1}{x} \implies du = (1 + \frac{1}{x^2}) dx$.
4. Use the standard integral: $\int \frac{du}{u^2 + a^2} = \frac{1}{a} \tan^{-1} \left( \frac{u}{a} \right) + C$.
Step 3: Detailed Explanation:
Let $I = \int \frac{x^{2}+1}{x^{4}+1} dx$.
Dividing both terms by $x^2$:
\[ I = \int \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2}} dx \]
We need to express $x^2 + \frac{1}{x^2}$ in terms of a substitution whose derivative is $(1 + \frac{1}{x^2})$.
Consider $(x - \frac{1}{x})^2 = x^2 + \frac{1}{x^2} - 2$.
This implies $x^2 + \frac{1}{x^2} = (x - \frac{1}{x})^2 + 2$.
Substituting this into our integral:
\[ I = \int \frac{(1 + \frac{1}{x^2}) dx}{(x - \frac{1}{x})^2 + 2} \]
Now, let $u = x - \frac{1}{x}$.
Differentiating both sides: $du = (1 + \frac{1}{x^2}) dx$.
The integral transforms into:
\[ I = \int \frac{du}{u^2 + (\sqrt{2})^2} \]
This is a standard form. Applying the formula:
\[ I = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{u}{\sqrt{2}} \right) + C \]
Now, substitute the value of $u$ back in terms of $x$:
$u = \frac{x^2 - 1}{x}$.
\[ I = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{x^2 - 1}{\sqrt{2}x} \right) + C \]
This matches Option (A).
Step 4: Final Answer:
The indefinite integral is $\frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{x^{2}-1}{\sqrt{2}x} \right) + C$.