Question

If $\int \frac{1+x^2}{1+x^4} d x=\frac{1}{\sqrt{2}} \tan ^{-1}\left[\frac{f(x)}{\sqrt{2}}\right]+c$, then $f(x)=$

• A. $x + \frac{1}{x}$
• B. $x - \frac{1}{x}$
• C. $x + \frac{2}{x}$
• D. $x - \frac{2}{x}$

Hint:

Whenever you see a derivative term matching $(1 + 1/x^2)dx$ in the numerator, your integration substitution variable must always be its matching anti-derivative pair: $(x - 1/x)$. This instantly pinpoints option (B) without expanding the full calculation!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem presents an integration identity where we need to find the functional form of $f(x)$ inside the inverse tangent function after evaluating the indefinite integral.

Step 2: Key Formula or Approach:
This is a standard algebraic fractional integration template. We can solve it by dividing both the numerator and the denominator by $x^2$ and then executing an algebraic substitution.

Step 3: Detailed Explanation:
Let the given integral be $I$: $$ I = \int \frac{1+x^2}{1+x^4} d x $$ Dividing the numerator and the denominator by $x^2$: $$ I = \int \frac{\frac{1}{x^2} + 1}{\frac{1}{x^2} + x^2} d x = \int \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2}} d x $$ Let's rewrite the denominator by completing the square in terms of a subtraction binomial: $$ x^2 + \frac{1}{x^2} = \left(x - \frac{1}{x}\right)^2 + 2 $$ Substituting this back into the denominator yields: $$ I = \int \frac{1 + \frac{1}{x^2}}{\left(x - \frac{1}{x}\right)^2 + 2} d x $$ Now, let's use the substitution method: $$ \text{Put } x - \frac{1}{x} = t $$ Differentiating both sides with respect to $x$: $$ \left(1 - \left(-\frac{1}{x^2}\right)\right) d x = d t \implies \left(1 + \frac{1}{x^2}\right) d x = d t $$ Our integral transforms cleanly to: $$ I = \int \frac{d t}{t^2 + 2} = \int \frac{d t}{t^2 + (\sqrt{2})^2} $$ Applying the standard integration rule $\int \frac{d x}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + c$: $$ I = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{t}{\sqrt{2}}\right) + c $$ Replacing $t$ with its original expression $x - \frac{1}{x}$: $$ I = \frac{1}{\sqrt{2}} \tan^{-1}\left[\frac{x - \frac{1}{x}}{\sqrt{2}}\right] + c $$ Comparing this result directly with the given formula reveals that $f(x) = x - \frac{1}{x}$.

Step 4: Final Answer:
The function $f(x)$ is equal to $x - \frac{1}{x}$, which corresponds to option (B).