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

To solve the integral:

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

we use substitution and trigonometric identities. Let's break it down step-by-step:

1. Recognize that the integral involves a trigonometric form. We can use the substitution: \(x = \tan \theta\). Therefore, \(dx = \sec^2 \theta \, d\theta\).
2. Substitute these into the integral: 

\[\int \frac{1 - \tan^2 \theta}{(1 + \tan^2 \theta)\sqrt{1 + \tan^4 \theta}} \sec^2 \theta \, d\theta\]

1. Use the trigonometric identities: \(1 + \tan^2 \theta = \sec^2 \theta\) and \(1 - \tan^2 \theta = \frac{1}{\sec^2 \theta} - \tan^2 \theta\). Simplifying the integral, we have: 

\[\int \frac{\sec^2 \theta - \tan^2 \theta}{\sec^2 \theta \cdot \sec \theta} \, d\theta\]

1. Simplifying further: 

\[\int \cos \theta \, d\theta = \sin \theta + C\]

1. Substitute back \(\tan \theta = x\), so \(\sin \theta = \frac{x}{\sqrt{1 + x^2}}\), giving: 

\[\int \frac{1 - x^2}{(1 + x^2)\sqrt{1 + x^4}} \, dx = \frac{1}{\sqrt{2}} \sin^{-1}\left\{\frac{\sqrt{2}x}{x^2+1}\right\} + C\]

Thus, the correct answer is:

\(\frac{1}{\sqrt{2}}\sin^{-1}\left\{\frac{\sqrt{2}x}{x^2+1}\right\} + C\)

This confirms the option: \(\frac{1}{\sqrt{2}}\sin^{-1}\left\{\frac{\sqrt{2}x}{x^2+1}\right\} + C\) is correct.