`int(1-x^(2))/((1+x^(2))sqrt(1+x^(4)))dx` is equal to

Options

A. `sqrt(2)sin^(-1){(sqrt(2)x)/(x^(2)+1)}+C`
B. `(1)/(sqrt(2))sin^(-1){(sqrt(2)x)/(x^(2)+1)}`
C. `(1)/(2)sin^(-1){(sqrt(2)x*)/(x^(2)+1)}+C`
D. none of these

Explanation

To solve the integral

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

we will follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the integral:

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

### Step 2: Simplify the Expression
We can divide the numerator and the denominator by \(x^2\):

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

This gives us:

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

### Step 3: Rewrite the Square Root
Now, we can simplify the square root in the denominator. We rewrite \(1 + \frac{1}{x^2}\) as:

\[ \sqrt{1 + \frac{1}{x^2}} = \sqrt{\frac{x^2 + 1}{x^2}} = \frac{\sqrt{x^2 + 1}}{x}. \]

Thus, the integral becomes:

\[ \int \frac{\frac{1}{x^2} - 1}{\frac{1}{x} + x \cdot \frac{\sqrt{x^2 + 1}}{x}} \, dx = \int \frac{\frac{1}{x^2} - 1}{\frac{1 + x \sqrt{x^2 + 1}}{x}} \, dx. \]

### Step 4: Substitute
Let us make the substitution:

\[ t = \frac{1}{x} + x. \]

Differentiating gives:

\[ dt = \left(-\frac{1}{x^2} + 1\right) dx. \]

Thus,

\[ dx = \frac{dt}{-\frac{1}{x^2} + 1}. \]

### Step 5: Substitute Back into the Integral
Substituting \(t\) into the integral, we have:

\[ \int \frac{-dt}{t \sqrt{t^2 - 2}}. \]

### Step 6: Use Integration Formula
We can use the integration formula:

\[ \int \frac{-dx}{x \sqrt{x^2 - a^2}} = \frac{1}{a} \cos^{-1} \left(\frac{x}{a}\right) + C. \]

Here, \(a = \sqrt{2}\), so we have:

\[ \frac{1}{\sqrt{2}} \cos^{-1} \left(\frac{t}{\sqrt{2}}\right) + C. \]

### Step 7: Substitute Back for \(t\)
Substituting back for \(t\):

\[ \frac{1}{\sqrt{2}} \cos^{-1} \left(\frac{\frac{1}{x} + x}{\sqrt{2}}\right) + C. \]

### Step 8: Final Simplification
This can be rewritten in terms of \(\sin^{-1}\):

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

Thus, the final answer is:

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