Mark M. answered 10/26/18
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
Let F(x) = 6 + integral (from c to x) [f(t) / t]dt = 2sqrt(x)
By the Fundamental Theorem of Calculus, F'(x) = f(x) / x = 1 / sqrt(x)
So, f(x) = x / sqrt(x).
Also, F(c) = 6 + integral (from c to c) [f(t)/t]dt = 2sqrt(c)
So, 6 + 0 = 2sqrt(c)
sqrt(c) = 3 Thus, c = 9.