evaluate the following integral or state that it diverges.\n int_{5}^{infty}\frac{1}{(x + 4)^{2}}dx \nselect…

Explanation:

Step1: Rewrite as a limit

The improper integral $\int_{5}^{\infty}\frac{1}{(x + 4)^2}dx=\lim_{b\rightarrow\infty}\int_{5}^{b}\frac{1}{(x + 4)^2}dx$.

Step2: Use substitution

Let $u=x + 4$, then $du=dx$. When $x = 5$, $u=9$; when $x = b$, $u=b + 4$. So $\lim_{b\rightarrow\infty}\int_{5}^{b}\frac{1}{(x + 4)^2}dx=\lim_{b\rightarrow\infty}\int_{9}^{b + 4}u^{-2}du$.

Step3: Integrate

The antiderivative of $u^{-2}$ is $-u^{-1}+C=-\frac{1}{u}+C$. Then $\lim_{b\rightarrow\infty}\int_{9}^{b + 4}u^{-2}du=\lim_{b\rightarrow\infty}\left[-\frac{1}{u}\right]_{9}^{b + 4}$.

Step4: Evaluate the limit

$\lim_{b\rightarrow\infty}\left(-\frac{1}{b + 4}+\frac{1}{9}\right)=0+\frac{1}{9}=\frac{1}{9}$.

Answer:

A. The improper integral converges and $\int_{5}^{\infty}\frac{1}{(x + 4)^2}dx=\frac{1}{9}$