Question

integration infinity to 3 2/x2-1 dx​

Answer 1

Answer: ln 2

Step-by-step explanation:

First step is to break this into partial fractions

$\frac{2}{x^2 - 1} = \frac{2}{(x+1)(x-1)} = \frac{1}{x-1} - \frac{1}{x+1}$

Next you can proceed to do integration

$\int\limits^\infty_3 {\frac{2}{x^2-1}} \, dx \\= \int\limits^\infty_3 {\frac{1}{x-1} - \frac{1}{x+1}}} \, dx\\= \left [\ln (x-1) - \ln(x+1)]^\infty _3\\= [\ln(\frac{x-1}{x+1})]^\infty_3\\= [\ln 1 - \ln \frac{1}{2}]\\= \ln 2$