Question
The integral in this exercise converges. Evaluate the integral without using a table. \( \int_{-\infty}^{\infty} \frac{2 x d x}{\left(x^{2}+3\right)^{2}} \) \( \int_{-\infty}^{\infty} \frac{2 x d x}{\left(x^{2}+3\right)^{2}}=\square \)
Ask by Chandler Cox. in the United States
Jan 24,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The integral evaluates to 0.
Solution
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:
\[
\int_{-\infty}^{\infty} \frac{2x \, dx}{\left(x^{2}+3\right)^{2}} = 0.
\]
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Simplify this solution Bonus Knowledge
To evaluate the integral \( \int_{-\infty}^{\infty} \frac{2 x d x}{\left(x^{2}+3\right)^{2}} \), we can utilize the symmetry of the integrand. Notice that \( \frac{2x}{(x^2 + 3)^2} \) is an odd function since \( f(-x) = -f(x) \). The integral of an odd function over a symmetric interval, like \(-\infty\) to \(\infty\), is 0. Therefore, we conclude: \[ \int_{-\infty}^{\infty} \frac{2 x d x}{\left(x^{2}+3\right)^{2}} = 0 \]
