The value of the integral `int _0^oo 1/(1+x^4)dx` is

Options

A. `(pi)/(2)`
B. `(pi)/(sqrt(2))`
C. `(pi)/(2sqrt(2))`
D. none of these

Answer

To solve the integral \( I = \int_0^\infty \frac{1}{1 + x^4} \, dx \), we will use a series of substitutions and properties of definite integrals.

### Step 1: Define the integral
Let
\[ I = \int_0^\infty \frac{1}{1 + x^4} \, dx \]

### Step 2: Split the integral
We can rewrite the integrand by adding and subtracting \( x^2 \) in the numerator:
\[ I = \int_0^\infty \frac{x^2 + 1 - x^2}{1 + x^4} \, dx = \int_0^\infty \frac{x^2}{1 + x^4} \, dx + \int_0^\infty \frac{1 - x^2}{1 + x^4} \, dx \]

### Step 3: Define the second integral
Let
\[ I_2 = \int_0^\infty \frac{1 - x^2}{1 + x^4} \, dx \]
Thus, we can express \( I \) as:
\[ I = \int_0^\infty \frac{x^2}{1 + x^4} \, dx + I_2 \]

### Step 4: Analyze \( I_2 \)
We can simplify \( I_2 \) by making the substitution \( x = \frac{1}{t} \). Then \( dx = -\frac{1}{t^2} dt \), and the limits change as follows:
- When \( x = 0 \), \( t \to \infty \)
- When \( x \to \infty \), \( t \to 0 \)

Thus,
\[ I_2 = \int_\infty^0 \frac{1 - \frac{1}{t^2}}{1 + \frac{1}{t^4}} \left(-\frac{1}{t^2}\right) dt = \int_0^\infty \frac{t^2 - 1}{t^4 + 1} dt \]

### Step 5: Combine integrals
Now we have:
\[ I_2 = -\int_0^\infty \frac{1}{t^4 + 1} dt + \int_0^\infty \frac{t^2}{t^4 + 1} dt \]
Notice that:
\[ I_2 = -I + I \]
This implies:
\[ I_2 = -I + I = 0 \]

### Step 6: Substitute back into the equation
Substituting \( I_2 = 0 \) back into our expression for \( I \):
\[ I = \int_0^\infty \frac{x^2}{1 + x^4} \, dx \]

### Step 7: Evaluate \( I \)
Now, we can evaluate \( I \) by using the symmetry of the integrand. Notice that:
\[ \int_0^\infty \frac{x^2}{1 + x^4} \, dx = \int_0^\infty \frac{1}{1 + x^4} \, dx \]
Thus, we can write:
\[ 2I = \int_0^\infty \frac{1 + x^2}{1 + x^4} \, dx \]

### Step 8: Final evaluation
Using the substitution \( x = \tan(\theta) \), we can evaluate the integral:
\[ \int_0^{\frac{\pi}{2}} \sec^2(\theta) \, d\theta = \left[ \tan(\theta) \right]_0^{\frac{\pi}{2}} = \frac{\pi}{2} \]
Thus,
\[ I = \frac{\pi}{4\sqrt{2}} \]

### Conclusion
The value of the integral \( \int_0^\infty \frac{1}{1 + x^4} \, dx \) is:
\[ \boxed{\frac{\pi}{2\sqrt{2}}} \]