The value of b > 3 for which 12 ∫ 3 b 1 x 2 - 1 x 2 - 4 d x = log e 49 40 , is equal to _____.

Given $12\times {\int }_3^b\frac{1dx}{\left(x^2-4\right)\left(x^2-1\right)}=\ln \left[\frac{49}{40}\right] \& 4 b>3$

$\Rightarrow$ By Using partial fraction

$\frac{1}{\left(x^2-4\right)\left(x^2-1\right)}=\frac{1}{3}\left(\frac{1}{x^2-4}-\frac{1}{x^2-1}\right)$

$\Rightarrow \frac{12}{3}{\int }_3^b\left[\frac{1}{x^2-4}-\frac{1}{x^2-1}\right]=\ln \left(\frac{49}{40}\right)$

$\Rightarrow 4{\left[\frac{1}{4}\log \left(\frac{x-2}{x+2}\right)-\frac{1}{2}\log \left(\frac{x-1}{x+1}\right)\right]}_3^b=\ln \left(\frac{49}{40}\right)$

$\Rightarrow \log \left(\frac{\left(b-2\right){\left(b+1\right)}^2}{\left(b+2\right){\left(b-1\right)}^2}\times \frac{5}{4}\right)=\log \frac{49}{40}$

$\Rightarrow \frac{\left(b-2\right){\left(b+1\right)}^2}{\left(b+2\right){\left(b-1\right)}^2}\times \frac{5}{4}=\frac{49}{40}$

$\Rightarrow \frac{\left(b-2\right){\left(b+1\right)}^2}{\left(b+2\right){\left(b-1\right)}^2}=\frac{49}{50} ...\left(\mathrm{i}\right)$

Comparing by hit and trial and taking ${\left(b+1\right)}^2=49\Rightarrow b+1=7\Rightarrow b=6$ 

Now putting $b=6$ in equation (i) to corsscheck,

we get $\frac{4\times 7^2}{8\times 5^2}=\frac{49}{50}$ which is true 

So $b=6$  is correct option,

Hence $b=6$