the following integral converges. evaluate the integral without using tables.\n int_{3}^{infty}…

the following integral converges. evaluate the integral without using tables.\n int_{3}^{infty} \frac{8}{x^{2}+5 x+6} d x \n int_{3}^{infty} \frac{8}{x^{2}+5 x+6} d x=square \text { (type an exact answer.) }

the following integral converges. evaluate the integral without using tables.\n int_{3}^{infty} \frac{8}{x^{2}+5 x+6} d x \n int_{3}^{infty} \frac{8}{x^{2}+5 x+6} d x=square \text { (type an exact answer.) }

Answer

Explanation:

Step1: Decompose the integrand

Decompose (\frac{8}{x^{2}+5x + 6}) using partial - fraction. Factor (x^{2}+5x + 6=(x + 2)(x+3)). Let (\frac{8}{(x + 2)(x + 3)}=\frac{A}{x+2}+\frac{B}{x + 3}). Then (8=A(x + 3)+B(x + 2)). Set (x=-3), we get (8=B(-3 + 2)), so (B=-8). Set (x=-2), we get (8=A(-2+3)), so (A = 8). Thus (\frac{8}{x^{2}+5x + 6}=\frac{8}{x + 2}-\frac{8}{x + 3}).

Step2: Calculate the improper integral

(\int_{3}^{\infty}\frac{8}{x^{2}+5x + 6}dx=\lim_{t\rightarrow\infty}\int_{3}^{t}(\frac{8}{x + 2}-\frac{8}{x + 3})dx). Integrate term - by - term: (\int(\frac{8}{x + 2}-\frac{8}{x + 3})dx=8\ln|x + 2|-8\ln|x + 3|+C=8\ln\frac{x + 2}{x + 3}+C). Then (\lim_{t\rightarrow\infty}\left[8\ln\frac{x + 2}{x + 3}\right]{3}^{t}=\lim{t\rightarrow\infty}(8\ln\frac{t + 2}{t + 3}-8\ln\frac{3 + 2}{3 + 3})). Since (\lim_{t\rightarrow\infty}\frac{t + 2}{t + 3}=\lim_{t\rightarrow\infty}\frac{1+\frac{2}{t}}{1+\frac{3}{t}} = 1) and (\ln1 = 0). (8\ln1-8\ln\frac{5}{6}=8\ln\frac{6}{5}).

Answer:

(8\ln\frac{6}{5})