The value of b > 3 for which 12 ∫ _ 3 ^ b 1 ( x ^ 2 - 1)( x ^ 2 -...

Given $12 \times {∫}_{3}^{b}\dfrac{1dx}{\left({x}^{2} - 4\right)\left({x}^{2} - 1\right)} = \ln \left[\dfrac{49}{40}\right] \text{ and } 4b > 3$

$⇒$ By Using partial fraction

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

$⇒ \dfrac{12}{3}{∫}_{3}^{b}\left[\dfrac{1}{{x}^{2} - 4} - \dfrac{1}{{x}^{2} - 1}\right] = \ln \left(\dfrac{49}{40}\right)$

$⇒ 4{\left[\dfrac{1}{4}\log \left(\dfrac{x - 2}{x + 2}\right) - \dfrac{1}{2}\log \left(\dfrac{x - 1}{x + 1}\right)\right]}_{3}^{b} = \ln \left(\dfrac{49}{40}\right)$

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

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

$⇒ \dfrac{\left(b - 2\right){\left(b + 1\right)}^{2}}{\left(b + 2\right){\left(b - 1\right)}^{2}} = \dfrac{49}{50} . . . \left(i\right)$

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

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

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

So $b = 6$ is correct option,

Hence $b = 6$