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 \( I = \int_{0}^{\infty} \frac{dx}{(1 + x^{2020})(1 + x^{2})} \) and find the value of \( 16k \) where \( I = k\pi \), we can follow these steps:

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

### Step 2: Substitute in the Integral
Substituting \( x = \tan \theta \) into the integral, we have:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sec^2 \theta \, d\theta}{(1 + \tan^{2020} \theta)(1 + \tan^2 \theta)} \]
Since \( 1 + \tan^2 \theta = \sec^2 \theta \), we can simplify:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{d\theta}{(1 + \tan^{2020} \theta)} \]

### Step 3: Use Symmetry Property of Integrals
We can use the property of integrals:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{d\theta}{(1 + \tan^{2020} \theta)} = \int_{0}^{\frac{\pi}{2}} \frac{d\theta}{(1 + \cot^{2020} \theta)} \]
Let’s denote the second integral as \( I' \):
\[ I' = \int_{0}^{\frac{\pi}{2}} \frac{d\theta}{(1 + \cot^{2020} \theta)} = \int_{0}^{\frac{\pi}{2}} \frac{\tan^{2020} \theta}{(1 + \tan^{2020} \theta)} \, d\theta \]

### Step 4: Combine the Integrals
Adding both integrals:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \left( \frac{1}{1 + \tan^{2020} \theta} + \frac{1}{1 + \cot^{2020} \theta} \right) d\theta \]
This simplifies to:
\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{1 + \tan^{2020} \theta}{(1 + \tan^{2020} \theta)(1 + \cot^{2020} \theta)} \, d\theta \]
Since \( \cot \theta = \frac{1}{\tan \theta} \), we can simplify further.

### Step 5: Evaluate the Integral
This integral evaluates to:
\[ 2I = \int_{0}^{\frac{\pi}{2}} d\theta = \frac{\pi}{2} \]
Thus, we find:
\[ I = \frac{\pi}{4} \]

### Step 6: Relate to \( k \)
Since \( I = k\pi \), we have:
\[ \frac{\pi}{4} = k\pi \implies k = \frac{1}{4} \]

### Step 7: Calculate \( 16k \)
Now, we calculate \( 16k \):
\[ 16k = 16 \times \frac{1}{4} = 4 \]

Thus, the final answer is:
\[ \boxed{4} \]