$$I = \int {{1 \over {({x^2} - 1)({x^2} - 4)}}dx = {1 \over 3}\int {\left( {{1 \over {{x^2} - 4}} - {1 \over {{x^2} - 1}}} \right)dx} } $$

$$ = {1 \over 3}\left( {{1 \over 4}\ln \left| {{{x - 2} \over {x + 2}}} \right| - {1 \over 2}\ln \left| {{{x - 1} \over {x + 1}}} \right|} \right) + C$$

$$12I = \ln \left| {{{x - 2} \over {x + 2}}} \right| + 2\ln \left| {{{x - 1} \over {x + 1}}} \right| + C$$

$$12\int\limits_3^b {{{dx} \over {({x^2} - 4)({x^2} - 1)}}} $$

$$ = \ln \left( {{{b - 2} \over {b + 2}}} \right) - 2\ln \left( {{{b - 1} \over {b + 1}}} \right) - \left( {\ln \left( {{1 \over 5}} \right) - 2\ln \left( {{1 \over 2}} \right)} \right)$$

$$ = \ln \left( {\left( {{{b - 2} \over {b + 2}}} \right)\,.\,{{{{(b + 1)}^2}} \over {{{(b - 1)}^2}}}} \right) - \left( {\ln {4 \over 5}} \right)$$

So, $${{49} \over {40}} = {{(b - 2)} \over {(b + 2)}}{{{{(b + 1)}^2}} \over {{{(b - 1)}^2}}}\,.\,{5 \over 4}$$

$$ \Rightarrow b = 6$$