Question

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

Answer 1

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