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: Divide top and bottom by $x^2$.
\[ \int\frac{1+x^2}{1+x^4}\,dx=\int\frac{\frac{1}{x^2}+1}{\frac{1}{x^2}+x^2}\,dx \]

Step 2: Rewrite the denominator.
Note $x^2+\frac{1}{x^2}=\left(x-\frac{1}{x}\right)^2+2$. The numerator is $1+\frac{1}{x^2}$.

Step 3: Substitute.
Let $t=x-\frac{1}{x}$, so $dt=\left(1+\frac{1}{x^2}\right)dx$. The integral turns into $\int\frac{dt}{t^2+2}$.

Step 4: Integrate and match.
\[ \frac{1}{\sqrt2}\tan^{-1}\left(\frac{t}{\sqrt2}\right)+c \] So $f(x)=t=x-\frac{1}{x}$. \[ \boxed{f(x)=x-\frac{1}{x}} \]