`int_(0)^(oo)(x^(2)+1)/(x^(4)+7x^(2)+1)dx=`

Answer

To solve the integral

\[ I = \int_{0}^{\infty} \frac{x^2 + 1}{x^4 + 7x^2 + 1} \, dx, \]

we will follow these steps:

### Step 1: Simplify the Integral
We start by rewriting the integral in a more manageable form. We can factor out \(x^2\) from both the numerator and the denominator:

\[ I = \int_{0}^{\infty} \frac{x^2(1 + \frac{1}{x^2})}{x^2(x^2 + 7 + \frac{1}{x^2})} \, dx. \]

This simplifies to:

\[ I = \int_{0}^{\infty} \frac{1 + \frac{1}{x^2}}{x^2 + 7 + \frac{1}{x^2}} \, dx. \]

### Step 2: Rewrite the Denominator
Next, we can rewrite the denominator:

\[ x^2 + 7 + \frac{1}{x^2} = \left(x - \frac{1}{x}\right)^2 + 9. \]

This gives us:

\[ I = \int_{0}^{\infty} \frac{1 + \frac{1}{x^2}}{\left(x - \frac{1}{x}\right)^2 + 9} \, dx. \]

### Step 3: Change of Variables
Now, we will perform a substitution. Let:

\[ t = x - \frac{1}{x}. \]

Differentiating gives us:

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

Thus, we can rewrite the integral as:

\[ I = \int_{-\infty}^{\infty} \frac{1}{t^2 + 9} \, dt. \]

### Step 4: Evaluate the Integral
The integral

\[ \int \frac{1}{t^2 + a^2} \, dt = \frac{1}{a} \tan^{-1}\left(\frac{t}{a}\right) + C. \]

In our case, \(a = 3\):

\[ I = \frac{1}{3} \left[ \tan^{-1}\left(\frac{t}{3}\right) \right]_{-\infty}^{\infty}. \]

### Step 5: Compute the Limits
Now we compute the limits:

- As \(t \to \infty\), \(\tan^{-1}\left(\frac{t}{3}\right) \to \frac{\pi}{2}\).
- As \(t \to -\infty\), \(\tan^{-1}\left(\frac{t}{3}\right) \to -\frac{\pi}{2}\).

Thus,

\[ I = \frac{1}{3} \left( \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) \right) = \frac{1}{3} \cdot \pi = \frac{\pi}{3}. \]

### Final Answer
Therefore, the value of the integral is:

\[ \boxed{\frac{\pi}{3}}. \]