We are asked to evaluate the integral $J = \int \frac{dx}{(x^2 + 1)^2}$.

1. Problem Description

We are asked to evaluate the integral $J = \int \frac{dx}{(x^2 + 1)^2}$.

2. Solution Steps

We will use the trigonometric substitution $x = \tan(\theta)$, which implies $dx = \sec^2(\theta) d\theta$.

Then $x^2 + 1 = \tan^2(\theta) + 1 = \sec^2(\theta)$.

Substituting into the integral, we have:

$J = \int \frac{\sec^2(\theta) d\theta}{(\sec^2(\theta))^2} = \int \frac{\sec^2(\theta)}{\sec^4(\theta)} d\theta = \int \frac{1}{\sec^2(\theta)} d\theta = \int \cos^2(\theta) d\theta$.

We use the identity $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$ to simplify the integral.

$J = \int \frac{1 + \cos(2\theta)}{2} d\theta = \frac{1}{2} \int (1 + \cos(2\theta)) d\theta = \frac{1}{2} \left( \theta + \frac{1}{2}\sin(2\theta) \right) + C$.

We can rewrite $\sin(2\theta)$ as $2\sin(\theta)\cos(\theta)$.

So, $J = \frac{1}{2} \theta + \frac{1}{2} \sin(\theta)\cos(\theta) + C$.

Since $x = \tan(\theta)$, we have $\theta = \arctan(x)$.

Also, $\tan(\theta) = \frac{x}{1}$. We can consider a right triangle where the opposite side is $x$ and the adjacent side is $1$. The hypotenuse is $\sqrt{x^2 + 1}$.

Then $\sin(\theta) = \frac{x}{\sqrt{x^2+1}}$ and $\cos(\theta) = \frac{1}{\sqrt{x^2+1}}$.

Therefore, $\sin(\theta)\cos(\theta) = \frac{x}{x^2 + 1}$.

Substituting these back into our expression for $J$, we get

$J = \frac{1}{2} \arctan(x) + \frac{1}{2} \left(\frac{x}{x^2 + 1} \right) + C = \frac{1}{2} \arctan(x) + \frac{x}{2(x^2 + 1)} + C$.

3. Final Answer

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