evaluate the following integral or state that it diverges. ∫ from -∞ to ∞ dx / (x² + 256) select the correct…

Explanation:

Step1: Rewrite the improper - integral

We know that $\int_{-\infty}^{\infty}\frac{dx}{x^{2}+256}=\int_{-\infty}^{0}\frac{dx}{x^{2}+256}+\int_{0}^{\infty}\frac{dx}{x^{2}+256}$. Let's first consider $\int\frac{dx}{x^{2}+256}$. We use the formula $\int\frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\arctan(\frac{x}{a})+C$. Here $a = 16$ since $x^{2}+256=x^{2}+16^{2}$, so $\int\frac{dx}{x^{2}+256}=\frac{1}{16}\arctan(\frac{x}{16})+C$.

Step2: Evaluate $\int_{0}^{\infty}\frac{dx}{x^{2}+256}$

$\lim_{b\rightarrow\infty}\int_{0}^{b}\frac{dx}{x^{2}+256}=\lim_{b\rightarrow\infty}\left[\frac{1}{16}\arctan(\frac{x}{16})\right]{0}^{b}=\lim{b\rightarrow\infty}\frac{1}{16}\arctan(\frac{b}{16})-\frac{1}{16}\arctan(0)$. Since $\lim_{u\rightarrow\infty}\arctan(u)=\frac{\pi}{2}$ and $\arctan(0) = 0$, we have $\lim_{b\rightarrow\infty}\frac{1}{16}\arctan(\frac{b}{16})=\frac{\pi}{32}$.

Step3: Evaluate $\int_{-\infty}^{0}\frac{dx}{x^{2}+256}$

Let $c\rightarrow-\infty$. Then $\int_{c}^{0}\frac{dx}{x^{2}+256}=\lim_{c\rightarrow-\infty}\left[\frac{1}{16}\arctan(\frac{x}{16})\right]{c}^{0}=\frac{1}{16}\arctan(0)-\lim{c\rightarrow-\infty}\frac{1}{16}\arctan(\frac{c}{16})$. Since $\lim_{u\rightarrow-\infty}\arctan(u)=-\frac{\pi}{2}$, we have $\lim_{c\rightarrow-\infty}\frac{1}{16}\arctan(\frac{c}{16})=-\frac{\pi}{32}$, so $\int_{-\infty}^{0}\frac{dx}{x^{2}+256}=\frac{\pi}{32}$.

Step4: Find the value of $\int_{-\infty}^{\infty}\frac{dx}{x^{2}+256}$

$\int_{-\infty}^{\infty}\frac{dx}{x^{2}+256}=\int_{-\infty}^{0}\frac{dx}{x^{2}+256}+\int_{0}^{\infty}\frac{dx}{x^{2}+256}=\frac{\pi}{32}+\frac{\pi}{32}=\frac{\pi}{16}$.

Answer:

A. The improper integral converges and $\int_{-\infty}^{\infty}\frac{dx}{x^{2}+256}=\frac{\pi}{16}$