Riemann integral equation

$\begin{aligned} I & = \int_a^{3a+1} \frac 1{x^2-x} dx \\ & = \int_a^{3a+1} \frac 1{x(x-1)} dx \\ & = \int_a^{3a+1} \left(\frac 1{x-1} - \frac 1x \right) dx \\ &= \ln (x-1) - \ln x \ \bigg|_a^{3a+1} \\ & = \ln \frac {3a}{3a+1} - \ln \frac {a-1}a \\ & = \ln \frac {3a^2}{(3a+1)(a-1)} \\ & = \ln \frac {27}{20} \end{aligned}$

Therefore, $a = \boxed 3$ .

Let us rewrite the above integrand as $\frac{1}{x^2-x} = \frac{1}{x-1} - \frac{1}{x}$ , which integrates according to:

$\int_{a}^{3a+1} \frac{1}{x-1} - \frac{1}{x} dx \Rightarrow \ln(x-1) - \ln(x)|_{a}^{3a+1} = \ln[\frac{3a^2}{(3a+1)(a-1)}] = \ln(\frac{27}{20}).$

Taking $\frac{3a^2}{3a^2-2a-1} = \frac{27}{20} \Rightarrow 0 = 21a^2 - 54a - 27 \Rightarrow 0 = 3(7a+3)(a-3) \Rightarrow a = -\frac{3}{7}, 3.$

Since $a > 1$ is a necessary & sufficient condition for the logarithmic expression above, the answer is $\boxed{a=3}.$

The value of the integral is $\ln |\frac{3a^2}{3a^2-2a-1}|$ .

Comparing with the R. H. S., we get $x=\boxed 3$ .