Let `u=int_0^oo (dx)/(x^4+7x^2+1` and `v=int_0^oo (x^2dx)/(x^4+7x^2+1)` then

Options

A. `pi//3`
B. `pi//6`
C. `pi//12`
D. `pi//9`

Explanation

To solve the given problem, we need to evaluate the integrals \( u \) and \( v \) defined as:

\[ u = \int_0^\infty \frac{dx}{x^4 + 7x^2 + 1} \]

\[ v = \int_0^\infty \frac{x^2 \, dx}{x^4 + 7x^2 + 1} \]

### Step 1: Combine the Integrals
First, we add the two integrals:

\[ u + v = \int_0^\infty \left( \frac{1}{x^4 + 7x^2 + 1} + \frac{x^2}{x^4 + 7x^2 + 1} \right) dx \]

This simplifies to:

\[ u + v = \int_0^\infty \frac{1 + x^2}{x^4 + 7x^2 + 1} \, dx \]

### Step 2: Factor the Denominator
Next, we can factor out \( x^2 \) from the denominator:

\[ u + v = \int_0^\infty \frac{1 + x^2}{x^2(x^2 + 7) + 1} \, dx \]

### Step 3: Substitute \( x = \frac{1}{t} \)
To simplify the integral, we can use the substitution \( x = \frac{1}{t} \). Then, \( dx = -\frac{1}{t^2} dt \), and the limits change from \( x = 0 \) to \( x = \infty \) to \( t = \infty \) to \( t = 0 \):

\[ u + v = \int_\infty^0 \frac{1 + \frac{1}{t^2}}{\frac{1}{t^4} + 7 \frac{1}{t^2} + 1} \left(-\frac{1}{t^2}\right) dt \]

Rearranging gives:

\[ u + v = \int_0^\infty \frac{t^2 + 1}{1 + 7t^2 + t^4} dt \]

### Step 4: Combine the Integrals Again
Now we have:

\[ u + v = \int_0^\infty \frac{1 + x^2}{x^4 + 7x^2 + 1} \, dx = \int_0^\infty \frac{1 + \frac{1}{t^2}}{\frac{1}{t^4} + 7 \frac{1}{t^2} + 1} \, dt \]

### Step 5: Evaluate the Integral
Next, we can evaluate \( u + v \) using the known result for integrals of this form. The integral evaluates to:

\[ u + v = \frac{\pi}{3} \]

### Step 6: Subtract the Integrals
Now, we subtract the two integrals:

\[ u - v = \int_0^\infty \frac{1 - x^2}{x^4 + 7x^2 + 1} \, dx \]

Using the same substitution \( x = \frac{1}{t} \):

\[ u - v = \int_0^\infty \frac{1 - \frac{1}{t^2}}{\frac{1}{t^4} + 7 \frac{1}{t^2} + 1} \left(-\frac{1}{t^2}\right) dt \]

This simplifies to:

\[ u - v = -\int_0^\infty \frac{t^2 - 1}{1 + 7t^2 + t^4} dt \]

### Step 7: Evaluate \( u - v \)
This integral can be evaluated, and it turns out that:

\[ u - v = 0 \]

### Step 8: Solve for \( u \) and \( v \)
From the equations \( u + v = \frac{\pi}{3} \) and \( u - v = 0 \), we can solve for \( u \) and \( v \):

Adding the two equations:

\[ 2u = \frac{\pi}{3} \implies u = \frac{\pi}{6} \]

Subtracting the second from the first:

\[ 2v = \frac{\pi}{3} \implies v = \frac{\pi}{6} \]

### Final Result
Thus, we have:

\[ u = \frac{\pi}{6}, \quad v = \frac{\pi}{6} \]