The problem asks to evaluate the integral: $\int \frac{x^{4037}}{(x^2 + 1)^{2020}} dx$

1. Problem Description

The problem asks to evaluate the integral:

$\int \frac{x^{4037}}{(x^2 + 1)^{2020}} dx$

2. Solution Steps

We can rewrite the exponent in the numerator as follows:

$x^{4037} = x^{4040 - 3} = x^{2(2020) - 3}$

Let $x = \tan{\theta}$. Then $dx = \sec^2{\theta} d\theta$. Also, $x^2 + 1 = \tan^2{\theta} + 1 = \sec^2{\theta}$.

The integral becomes:

$\int \frac{(\tan{\theta})^{4037}}{(\sec^2{\theta})^{2020}} \sec^2{\theta} d\theta = \int \frac{\tan^{4037}{\theta}}{\sec^{4040}{\theta}} \sec^2{\theta} d\theta = \int \frac{\tan^{4037}{\theta}}{\sec^{4038}{\theta}} d\theta$

$= \int \frac{\sin^{4037}{\theta}}{\cos^{4037}{\theta}} \cos^{4038}{\theta} d\theta = \int \sin^{4037}{\theta} \cos{\theta} d\theta$

Let $u = \sin{\theta}$, then $du = \cos{\theta} d\theta$.

Then the integral becomes $\int u^{4037} du = \frac{u^{4038}}{4038} + C = \frac{\sin^{4038}{\theta}}{4038} + C$.

Since $x = \tan{\theta}$, we have $\sin{\theta} = \frac{x}{\sqrt{x^2 + 1}}$.

Therefore,

$\frac{\sin^{4038}{\theta}}{4038} + C = \frac{1}{4038} \left(\frac{x}{\sqrt{x^2 + 1}}\right)^{4038} + C = \frac{x^{4038}}{4038 (x^2 + 1)^{2019}} + C$

3. Final Answer

$\frac{x^{4038}}{4038(x^2+1)^{2019}} + C$