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

Options

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

Answer

To solve the integral \( I = \int \frac{1 + x^2}{x \sqrt{1 + x^4}} \, dx \), we can follow these steps:

### Step 1: Simplify the integrand
We can rewrite the integrand by factoring out \( x^2 \) from the square root in the denominator:

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

This simplifies to:

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

### Step 2: Rewrite the integrand
Now, we can express the integrand as:

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

### Step 3: Change of variable
Let’s consider the substitution \( t = x - \frac{1}{x} \). Then, we compute the differential:

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

This means \( dx = \frac{dt}{1 + \frac{1}{x^2}} \).

### Step 4: Substitute in the integral
Substituting \( t \) into the integral, we have:

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

### Step 5: Solve the integral
The integral \( \int \frac{dt}{\sqrt{t^2 + 2}} \) can be solved using the standard integral formula:

\[ \int \frac{dx}{\sqrt{x^2 + a^2}} = \ln |x + \sqrt{x^2 + a^2}| + C \]

Thus, we have:

\[ I = \ln \left| t + \sqrt{t^2 + 2} \right| + C \]

### Step 6: Substitute back for \( t \)
Now, substituting back for \( t \):

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

### Final Answer
Thus, the final answer is:

\[ I = \ln \left| x - \frac{1}{x} + \sqrt{x^2 - 2 + \frac{1}{x^2} + 2} \right| + C \]