If ` I_(1)=int_(0)^(oo) (dx)/(1+x^(4))dx and I_(2)=underset(0)overset(oo)int x^(2)(dx)/(1+x^(4))"then"n (I_(1))/(I_(2))=`

Options

A. 1
B. 2
C. `1//2`
D. none of these

Answer

To solve the problem, we need to evaluate the integrals \( I_1 \) and \( I_2 \) given by:

\[ I_1 = \int_0^{\infty} \frac{dx}{1 + x^4} \]
\[ I_2 = \int_0^{\infty} \frac{x^2 \, dx}{1 + x^4} \]

Then, we will find the ratio \( \frac{I_1}{I_2} \).

### Step 1: Evaluate \( I_1 \)

We start with the integral \( I_1 \):

\[ I_1 = \int_0^{\infty} \frac{dx}{1 + x^4} \]

To evaluate this integral, we use the substitution \( x = \frac{1}{t} \). Then, \( dx = -\frac{1}{t^2} dt \). The limits change as follows: when \( x = 0 \), \( t \to \infty \) and when \( x \to \infty \), \( t = 0 \).

Substituting these into the integral gives:

\[ I_1 = \int_{\infty}^{0} \frac{-\frac{1}{t^2}}{1 + \left(\frac{1}{t}\right)^4} dt = \int_{0}^{\infty} \frac{1/t^2}{1 + \frac{1}{t^4}} dt \]

Rearranging the denominator:

\[ = \int_{0}^{\infty} \frac{1/t^2}{\frac{t^4 + 1}{t^4}} dt = \int_{0}^{\infty} \frac{t^4}{t^2(t^4 + 1)} dt = \int_{0}^{\infty} \frac{t^2}{1 + t^4} dt \]

Thus, we have:

\[ I_1 = \int_{0}^{\infty} \frac{t^2}{1 + t^4} dt \]

### Step 2: Evaluate \( I_2 \)

Now we evaluate \( I_2 \):

\[ I_2 = \int_0^{\infty} \frac{x^2 \, dx}{1 + x^4} \]

Notice that we have already derived that:

\[ I_1 = \int_0^{\infty} \frac{x^2}{1 + x^4} dx \]

Thus, we can conclude:

\[ I_1 = I_2 \]

### Step 3: Find the ratio \( \frac{I_1}{I_2} \)

Now, we find the ratio:

\[ \frac{I_1}{I_2} = \frac{I_1}{I_1} = 1 \]

### Final Answer

Thus, the final answer is:

\[ \frac{I_1}{I_2} = 1 \]