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 \( \int \frac{1+x^2}{1+x^4} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{1+x^2}{1+x^4} \, dx \]
We can separate the fraction:
\[ I = \int \left( \frac{1}{1+x^4} + \frac{x^2}{1+x^4} \right) \, dx \]

### Step 2: Simplify the Second Term
The term \( \frac{x^2}{1+x^4} \) can be rewritten:
\[ \frac{x^2}{1+x^4} = \frac{x^2}{(x^2)^2 + 1} = \frac{x^2}{x^4 + 1} \]

### Step 3: Use a Substitution
Let \( x^2 = t \), then \( dx = \frac{1}{2\sqrt{t}} \, dt \). The integral becomes:
\[ I = \int \left( \frac{1}{1+t^2} + \frac{t}{1+t^2} \right) \frac{1}{2\sqrt{t}} \, dt \]

### Step 4: Split the Integral
Now we can split the integral:
\[ I = \frac{1}{2} \int \frac{1}{1+t^2} \cdot \frac{1}{\sqrt{t}} \, dt + \frac{1}{2} \int \frac{t}{1+t^2} \cdot \frac{1}{\sqrt{t}} \, dt \]

### Step 5: Solve Each Integral
1. The first integral \( \int \frac{1}{1+t^2} \cdot \frac{1}{\sqrt{t}} \, dt \) can be solved using the arctangent formula:
\[ \int \frac{1}{1+t^2} \, dt = \tan^{-1}(t) \]
2. The second integral \( \int \frac{t}{1+t^2} \cdot \frac{1}{\sqrt{t}} \, dt \) can be simplified and solved using substitution.

### Step 6: Combine Results
After solving both integrals, we combine the results and substitute back \( t = x^2 \).

### Final Result
The final result will be:
\[ I = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x^2 - 1}{x}\right) + C \]