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

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

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

Answer

Explanation:

Step1: Decompose the integrand

First, factor the denominator (x^{2}+5x + 6=(x + 2)(x+3)). Use partial - fraction decomposition: (\frac{2}{x^{2}+5x + 6}=\frac{2}{(x + 2)(x + 3)}=\frac{A}{x+2}+\frac{B}{x + 3}). Multiply through by ((x + 2)(x + 3)) to get (2=A(x + 3)+B(x + 2)). Let (x=-3), then (2=B(-3 + 2)), so (B=-2). Let (x=-2), then (2=A(-2 + 3)), so (A = 2). So (\frac{2}{x^{2}+5x + 6}=\frac{2}{x + 2}-\frac{2}{x + 3}).

Step2: Evaluate the improper integral

(\int_{6}^{\infty}\frac{2}{x^{2}+5x + 6}dx=\lim_{t\rightarrow\infty}\int_{6}^{t}(\frac{2}{x + 2}-\frac{2}{x + 3})dx). Integrate term - by - term: (\int(\frac{2}{x + 2}-\frac{2}{x + 3})dx=2\ln|x + 2|-2\ln|x + 3|+C=2\ln\frac{x + 2}{x + 3}+C).

Step3: Apply the limits

(\lim_{t\rightarrow\infty}\left[2\ln\frac{x + 2}{x + 3}\right]{6}^{t}=\lim{t\rightarrow\infty}(2\ln\frac{t + 2}{t + 3}-2\ln\frac{6+2}{6 + 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). (2\ln\frac{8}{9}=2(\ln8-\ln9)=2(3\ln2 - 2\ln3)=6\ln2-4\ln3=\ln2^{6}-\ln3^{4}=\ln\frac{64}{81}).

Answer:

(\ln\frac{64}{81})