`T h e v a l u eofint_0^oo(dx)/(1+x^4)i s` (a) `s a m ea st h a tofint_0^oo(x^2+1dx)/(1+x)` (b) `pi/(2sqrt(2))` (c) `s a m ea st h a tofint_0^oo(x^2+1dx)/(1+x^4)` (d) `pi/(sqrt(2))`

Options

A. same as that of `int_(0)^(oo) (x^(2)+1dx)/(1+x^(4))`
B. `(pi)/(2sqrt(2))`
C. same as that of `int_(0)^(oo) (x^(2)dx)/(1+x^(4))`
D. `(pi)/(sqrt(2))`

Explanation

To solve the integral \( I = \int_0^\infty \frac{dx}{1+x^4} \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int_0^\infty \frac{dx}{1+x^4} \]

### Step 2: Use a Substitution
To simplify the integral, we can use the substitution \( x^2 = t \), which gives \( dx = \frac{1}{2\sqrt{t}} dt \). The limits change as follows: when \( x = 0 \), \( t = 0 \); and when \( x \to \infty \), \( t \to \infty \). Thus, we have:
\[ I = \int_0^\infty \frac{1}{1+t^2} \cdot \frac{1}{2\sqrt{t}} dt \]

### Step 3: Split the Integral
Next, we can express the integrand in a more manageable form:
\[ I = \frac{1}{2} \int_0^\infty \frac{1}{1+t^2} dt \]

### Step 4: Evaluate the Integral
The integral \( \int_0^\infty \frac{1}{1+t^2} dt \) is a standard integral that evaluates to \( \frac{\pi}{2} \). Therefore:
\[ I = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4} \]

### Step 5: Final Result
Thus, the value of the integral \( I \) is:
\[ I = \frac{\pi}{4} \]

### Step 6: Compare with Given Options
Now, we need to compare this result with the options given in the question. The closest match is:
- (b) \( \frac{\pi}{2\sqrt{2}} \)

### Conclusion
The value of the integral \( \int_0^\infty \frac{dx}{1+x^4} \) is \( \frac{\pi}{4} \), which corresponds to option (b).