indefinite integral?

Question

1/(3x-1)sqaure root dx

Answer 1

If you mean the integral of (square root of (1/(3x-1)) dx,
then you can integrate this by noting that if
f(x) = (3x-1)^1/2 (that ^ means to the power 1/2), then
f'(x) = 1/2(3x-1)^(-1/2) * 3, which is almost the integrand.
In fact, multiplying f(x) by 2/3 will do the trick:

f(x) = 2/3*sqrt(3x-1) is the answer!

Answer 2

∫1/√(3x-1) dx
= ∫(3x-1)^(-1/2) dx
=(1/3)∫(3x-1)^(-1/2) d(3x-1) ( mental substitution)
= (2/3)√(3x-1)