Question

The value of $b>3$ for which $12\underset{3}{\overset{b}{\int }}\frac{1}{\left(x^2-1\right)\left(x^2-4\right)}dx={\log }_e\left(\frac{49}{40}\right)$, is equal to

Answer 1

Answer: 6

$\frac{12}{3}\left[{\int }_3^b\left(\frac{1}{x^2-4}-\frac{1}{x^2-1}\right)dx\right]=\log \frac{49}{40}$
$\frac{12}{3}\cdot {\left[\frac{1}{4}\ln \left|\frac{x-2}{x+2}\right|-\frac{1}{2}\ln \left|\frac{x-1}{x+1}\right|\right]}_3^b=\log \frac{49}{40}$
$\ln \frac{(b-2)(b+1{)}^2}{(b+2)(b-1{)}^2}=\ln \frac{49}{50}$
$b=6$