Question
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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 Fleming Hilton. in the United States
Jan 24,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The integral evaluates to 0.

Solution

The integral to evaluate is: \[ \int_{-\infty}^{\infty} \frac{2x}{(x^2 + 3)^2} \, dx \] Let's analyze the integrand: 1. **Symmetry Consideration**: - The numerator \(2x\) is an **odd function** because \(2(-x) = -2x\). - The denominator \((x^2 + 3)^2\) is an **even function** because \((-x)^2 = x^2\). Therefore, the entire integrand \(\frac{2x}{(x^2 + 3)^2}\) is an **odd function**. 2. **Integral of an Odd Function Over Symmetric Limits**: - The integral of an odd function over symmetric limits \([-a, a]\) is zero because the contributions from \(x\) and \(-x\) cancel each other out. Hence, \[ \int_{-\infty}^{\infty} \frac{2x}{(x^2 + 3)^2} \, dx = 0 \] **Answer:** 0

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Beyond the Answer

To evaluate the integral \( \int_{-\infty}^{\infty} \frac{2 x \, dx}{(x^{2}+3)^{2}} \), we can notice that the integrand is an odd function. An odd function has the property that \( f(-x) = -f(x) \). Since the limits of integration are symmetric about the origin, the integral of an odd function over a symmetric interval around zero evaluates to zero. Thus, \[ \int_{-\infty}^{\infty} \frac{2 x \, dx}{(x^{2}+3)^{2}} = 0. \] Remember, integrals can lead to interesting symmetries in mathematics! This property can often help simplify problems without heavy lifting and heavy calculations. If you’re ever puzzled, take a moment to consider whether the function you're dealing with might have a useful symmetry or property to exploit. A seemingly challenging integral might turn into an effortless exercise simply because it’s odd or even!

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