I am not sure about how to exploit the a + bu idea. However, the standard approach with this type of integrand is the method of partial fractions.
1/(x (6+ 5x) can be decomposed as (1/6) /x - (5/6) /(5 x + 6)
Since both x and 5 x + 6 are positive on the interval [1,5] the anti derivative is
(1/6) ln(x) - (1/6) ln(5 x +6) = (1/6) ln( x /( 5 x + 6))
So the definite integral is (1/6) [ ln(5/31) - ln( 1/11) ] = 0.09556