Question

Evaluate the integral: 
\[ \int \frac{1 - x^2}{(1 + x^2)\sqrt{1 + x^4}} \, dx \]

• A. \(\sqrt{2}\sin^{-1}\left\{\frac{\sqrt{2}x}{x^2+1}\right\} + C\)
• B. \(\frac{1}{\sqrt{2}}\sin^{-1}\left\{\frac{\sqrt{2}x}{x^2+1}\right\} + C\)
• C. \(\frac{1}{2}\sin^{-1}\left\{\frac{\sqrt{2}x}{x^2+1}\right\} + C\)
• D. None of these

Hint:

When you see \(1+x^4\) in the denominator, try dividing by \(x^2\) and substitute \(t = x + 1/x\).

Answer 1

The Correct Option is B

Step 1: Concept:
Divide numerator and denominator by \(x^2\) and use substitution: \[ t = x + \frac{1}{x} \]
Step 2: Detailed Explanation:
Given, \[ I = \int \frac{1 - x^2}{(1 + x^2)\sqrt{1 + x^4}} \, dx \] Divide numerator and denominator by \(x^2\): \[ I = \int \frac{\frac{1}{x^2} - 1}{\left(x + \frac{1}{x}\right)\sqrt{x^2 + \frac{1}{x^2}}} \, dx \]
Let, \[ t = x + \frac{1}{x} \Rightarrow dt = \left(1 - \frac{1}{x^2}\right) dx = -\left(\frac{1}{x^2} - 1\right) dx \]
Also, \[ x^2 + \frac{1}{x^2} = t^2 - 2 \]
Substitute: \[ I = -\int \frac{dt}{t\sqrt{t^2 - 2}} \]
Using standard integral: \[ \int \frac{dt}{t\sqrt{t^2 - a^2}} = \frac{1}{a} \sin^{-1}\left(\frac{a}{t}\right) + C \] Here \( a = \sqrt{2} \), so: \[ I = \frac{1}{\sqrt{2}} \sin^{-1}\left(\frac{\sqrt{2}}{t}\right) + C \]
Substitute back \( t = x + \frac{1}{x} \): \[ I = \frac{1}{\sqrt{2}} \sin^{-1}\left(\frac{\sqrt{2}}{x + \frac{1}{x}}\right) + C \] Simplify: \[ \frac{\sqrt{2}}{x + \frac{1}{x}} = \frac{\sqrt{2}x}{x^2 + 1} \]
Thus, \[ I = \frac{1}{\sqrt{2}} \sin^{-1}\left(\frac{\sqrt{2}x}{x^2 + 1}\right) + C \]
Step 3: Final Answer:
\[ \boxed{ \frac{1}{\sqrt{2}} \sin^{-1}\left(\frac{\sqrt{2}x}{x^2 + 1}\right) + C } \]