The value of the integral `int(1+x^(2))/(1+x^(4))dx` is equal to

Options

A. `tan^(-1)x^(2)+C`
B. `(1)/(sqrt(2))tan^(-1)((x^(2)-1)/(sqrt(2)x))`
C. `(1)/(2sqrt(2))log((x^(2)+sqrt(2)x+1)/(x^(2)-sqrt(2)x+1))+C`
D. none of these

Answer

To solve the integral \( I = \int \frac{1 + x^2}{1 + x^4} \, dx \), we will follow a step-by-step approach.

### Step 1: Simplify the Integral
We start by dividing the numerator and the denominator by \( x^2 \):

\[ I = \int \frac{1 + x^2}{1 + x^4} \, dx = \int \frac{\frac{1}{x^2} + 1}{\frac{1}{x^2} + x^2} \, dx \]

### Step 2: Rewrite the Integral
Now we can rewrite the integral as follows:

\[ I = \int \frac{1 + \frac{1}{x^2}}{\frac{1}{x^2} + x^2} \, dx \]

### Step 3: Separate the Terms
Next, we can separate the terms in the integral:

\[ I = \int \left( \frac{1}{x^2 + 1} + \frac{1}{x^2} \right) \, dx \]

### Step 4: Use a Substitution
To solve the integral, we will make a substitution. Let:

\[ t = x - \frac{1}{x} \]

Differentiating both sides gives:

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

This means:

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

### Step 5: Rewrite the Integral in Terms of \( t \)
Now substituting \( dt \) into the integral, we have:

\[ I = \int \frac{dt}{t^2 + 2} \]

### Step 6: Solve the Integral
This integral can be solved using the formula for the integral of the form \( \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \):

\[ I = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{t}{\sqrt{2}} \right) + C \]

### Step 7: Substitute Back for \( t \)
Now we substitute back for \( t \):

\[ I = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{x - \frac{1}{x}}{\sqrt{2}} \right) + C \]

### Final Answer
Thus, the value of the integral is:

\[ I = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{x^2 - 1}{\sqrt{2} x} \right) + C \]