The integral in this exercise converges. Evaluate the integral without using a table. $$ \int_{-\infty}^{\infty} \frac{2 x d x}{\left(x^{2}+3\right)^{2}} $$ $$ \int_{-\infty}^{\infty} \frac{2 x d x}{\left(x^{2}+3\right)^{2}}=\square $$

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To evaluate the integral:

$$
\int_{-\infty}^{\infty} \frac{2x \, dx}{\left(x^{2}+3\right)^{2}},
$$

let's analyze the integrand:

$$
f(x) = \frac{2x}{\left(x^{2}+3\right)^{2}}.
$$

Notice that $$ f(x) $$ is an **odd function** because:

$$
f(-x) = \frac{2(-x)}{\left((-x)^{2}+3\right)^{2}} = -\frac{2x}{\left(x^{2}+3\right)^{2}} = -f(x).
$$

When integrating an odd function over a symmetric interval around zero (from $$-\infty$$ to $$+\infty$$), the positive and negative contributions cancel each other out, provided the integral converges. Since the problem states that the integral converges, we can conclude:

$$
\int_{-\infty}^{\infty} \frac{2x \, dx}{\left(x^{2}+3\right)^{2}} = 0.
$$

**Answer:** 0
The integral evaluates to 0.

The integral in this exercise converges. Evaluate the integral without using a table. \( \int_{-\infty}^{\infty} \frac{2 x d x}{\left(x^{2}+3\right)^{2}} \) \( \int_{-\infty}^{\infty} \frac{2 x d x}{\left(x^{2}+3\right)^{2}}=\square \)

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