the integral in this exercise converges. evaluate the integral without using a table\n int_{-infty}^{infty}…

Explanation:

Step1: Use substitution

Let (u = x^{2}+2), then (du=2x dx), and (4x dx = 2du). When (x =-\infty), (u=\infty); when (x=\infty), (u = \infty). The integral (\int_{-\infty}^{\infty}\frac{4x dx}{(x^{2}+2)^{5}}) becomes (2\int_{\infty}^{\infty}\frac{du}{u^{5}}).

Step2: Apply the power - rule for integration

The power - rule for integration is (\int u^{n}du=\frac{u^{n + 1}}{n+1}+C) ((n\neq - 1)). For (n=-5), (\int\frac{du}{u^{5}}=\int u^{-5}du=\frac{u^{-5 + 1}}{-5+1}+C=-\frac{1}{4u^{4}}+C). Then (2\int_{\infty}^{\infty}\frac{du}{u^{5}}=2\left[-\frac{1}{4u^{4}}\right]_{\infty}^{\infty}).

Step3: Evaluate the definite integral

(2\left(-\frac{1}{4u^{4}}\big|_{\infty}^{\infty}\right)=2\left[\left(-\frac{1}{4(\infty)^{4}}\right)-\left(-\frac{1}{4(\infty)^{4}}\right)\right]=0).

Answer:

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