the following integral converges. evaluate the integral without using tables.\n int_{6}^{infty}…

Explanation:

Step1: Decompose the integrand

First, factor the denominator (x^{2}+5x + 6=(x + 2)(x+3)). Use partial - fraction decomposition: (\frac{5}{(x + 2)(x + 3)}=\frac{A}{x+2}+\frac{B}{x + 3}). [ \begin{align*} 5&=A(x + 3)+B(x + 2)\ \end{align*} ] Let (x=-3), then (5=B(-3 + 2)), so (B=-5). Let (x=-2), then (5=A(-2+3)), so (A = 5). So (\frac{5}{x^{2}+5x + 6}=\frac{5}{x + 2}-\frac{5}{x + 3}).

Step2: Evaluate the improper integral

The improper integral (\int_{6}^{\infty}\frac{5}{x^{2}+5x + 6}dx=\lim_{b\rightarrow\infty}\int_{6}^{b}(\frac{5}{x + 2}-\frac{5}{x + 3})dx). Integrate term - by - term: (\int(\frac{5}{x + 2}-\frac{5}{x + 3})dx=5\ln|x + 2|-5\ln|x + 3|+C=5\ln\frac{x + 2}{x + 3}+C). Then (\lim_{b\rightarrow\infty}\left[5\ln\frac{x + 2}{x + 3}\right]{6}^{b}). [ \begin{align*} \lim{b\rightarrow\infty}\left(5\ln\frac{b + 2}{b + 3}-5\ln\frac{6+2}{6 + 3}\right)&=5\lim_{b\rightarrow\infty}\ln\frac{b + 2}{b + 3}-5\ln\frac{8}{9}\ \end{align*} ] Since (\lim_{b\rightarrow\infty}\frac{b + 2}{b + 3}=\lim_{b\rightarrow\infty}\frac{1+\frac{2}{b}}{1+\frac{3}{b}} = 1) and (\ln1 = 0).

Answer:

(5\ln\frac{9}{8})