The value fo the integral `I=int_(0)^(oo)(dx)/((1+x^(2020))(1+x^(2)))` is equal to `kpi`, then the value of 16k is equal to

Answer

To solve the integral

\[ I = \int_0^{\infty} \frac{dx}{(1+x^{2020})(1+x^2)}, \]

we will use a substitution and properties of integrals to evaluate it step by step.

### Step 1: Substitution
Let \( x = \tan \theta \). Then, \( dx = \sec^2 \theta \, d\theta \). The limits change as follows:
- When \( x = 0 \), \( \theta = 0 \).
- When \( x \to \infty \), \( \theta \to \frac{\pi}{2} \).

Thus, the integral becomes:

\[ I = \int_0^{\frac{\pi}{2}} \frac{\sec^2 \theta \, d\theta}{(1+\tan^{2020} \theta)(1+\tan^2 \theta)}. \]

### Step 2: Simplifying the Denominator
Recall that \( 1 + \tan^2 \theta = \sec^2 \theta \). Therefore, we can rewrite the integral as:

\[ I = \int_0^{\frac{\pi}{2}} \frac{\sec^2 \theta \, d\theta}{(1+\tan^{2020} \theta)\sec^2 \theta} = \int_0^{\frac{\pi}{2}} \frac{d\theta}{1+\tan^{2020} \theta}. \]

### Step 3: Using the Property of Integrals
Now we will use the property of integrals:

\[ \int_0^{\frac{\pi}{2}} f(\theta) \, d\theta = \int_0^{\frac{\pi}{2}} f\left(\frac{\pi}{2} - \theta\right) \, d\theta. \]

For our case, let \( f(\theta) = \frac{1}{1+\tan^{2020} \theta} \). Then,

\[ \int_0^{\frac{\pi}{2}} f\left(\frac{\pi}{2} - \theta\right) \, d\theta = \int_0^{\frac{\pi}{2}} \frac{1}{1+\cot^{2020} \theta} \, d\theta. \]

### Step 4: Simplifying Further
Using the identity \( \cot \theta = \frac{1}{\tan \theta} \), we can rewrite the integral:

\[ \int_0^{\frac{\pi}{2}} \frac{1}{1+\cot^{2020} \theta} \, d\theta = \int_0^{\frac{\pi}{2}} \frac{\tan^{2020} \theta}{\tan^{2020} \theta + 1} \, d\theta. \]

### Step 5: Adding the Two Integrals
Now we have:

\[ 2I = \int_0^{\frac{\pi}{2}} \left( \frac{1}{1+\tan^{2020} \theta} + \frac{\tan^{2020} \theta}{\tan^{2020} \theta + 1} \right) d\theta = \int_0^{\frac{\pi}{2}} 1 \, d\theta = \frac{\pi}{2}. \]

### Step 6: Solving for \( I \)
Thus,

\[ 2I = \frac{\pi}{2} \implies I = \frac{\pi}{4}. \]

### Step 7: Finding \( k \)
Given that \( I = k\pi \), we have:

\[ k\pi = \frac{\pi}{4} \implies k = \frac{1}{4}. \]

### Step 8: Finding \( 16k \)
Finally, we compute:

\[ 16k = 16 \times \frac{1}{4} = 4. \]

Thus, the value of \( 16k \) is

\[ \boxed{4}. \]