The value of $$b > 3$$ for which $$12\int_3^b \frac{1}{(x^2 - 1)(x^2 - 4)} dx = \log_e\frac{49}{40}$$, is equal to ______.

We need to find $$b > 3$$ such that $$12\int_3^b \frac{1}{(x^2-1)(x^2-4)}\,dx = \log_e\frac{49}{40}$$.

Since $$\frac{1}{(x^2-1)(x^2-4)} = \frac{1}{3}\left(\frac{1}{x^2-4} - \frac{1}{x^2-1}\right)$$ and using $$\frac{1}{x^2-a^2} = \frac{1}{2a}\left(\frac{1}{x-a} - \frac{1}{x+a}\right)$$ gives $$\frac{1}{x^2-1} = \frac{1}{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right)$$ and $$\frac{1}{x^2-4} = \frac{1}{4}\left(\frac{1}{x-2} - \frac{1}{x+2}\right)$$.

Substituting into the integral yields
$$12 \cdot \frac{1}{3}\int_3^b \left[\frac{1}{4}\left(\frac{1}{x-2} - \frac{1}{x+2}\right) - \frac{1}{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right)\right]dx$$which simplifies to$$4\left[\frac{1}{4}\ln\frac{x-2}{x+2} - \frac{1}{2}\ln\frac{x-1}{x+1}\right]_3^b = \left[\ln\frac{x-2}{x+2} - 2\ln\frac{x-1}{x+1}\right]_3^b = \left[\ln\frac{x-2}{x+2} - \ln\left(\frac{x-1}{x+1}\right)^2\right]_3^b = \left[\ln\frac{(x-2)(x+1)^2}{(x+2)(x-1)^2}\right]_3^b$$.

Evaluating at the limits, one finds at $$x=3$$ that $$\frac{(1)(16)}{(5)(4)} = \frac{4}{5}$$ and at $$x=b$$ that $$\frac{(b-2)(b+1)^2}{(b+2)(b-1)^2}$$. Therefore the value of the integral is $$\ln\frac{(b-2)(b+1)^2}{(b+2)(b-1)^2} - \ln\frac{4}{5}$$ which must equal $$\ln\frac{49}{40}$$. Hence $$\ln\frac{(b-2)(b+1)^2}{(b+2)(b-1)^2} = \ln\frac{49}{40} + \ln\frac{4}{5} = \ln\frac{196}{200} = \ln\frac{49}{50}$$.

It follows that $$\frac{(b-2)(b+1)^2}{(b+2)(b-1)^2} = \frac{49}{50}$$. Trying $$b = 6$$ gives $$\frac{(4)(49)}{(8)(25)} = \frac{196}{200} = \frac{49}{50}$$, so the required value is $$\boxed{6}$$.