No.
To make it more clear for you, sub u = 6x-7, du = 6 dx. Therefore:
Well, you would start by letting u = 6x - 7
Take the derivative of both sides to get:
du = 6dx; Solve for dx to get: dx = du/6. Now substitute this into the original integral to get:
∫[(u)^(1/2)]du/6
Note, u^1/2 is the same as sqrt(u)
Now, because 1/6 is a constant you can put it in front of the integral.
(1/6)∫[u^(1/2)]du
By using the reverse power rule, the integration would result in:
(1/6)*(2/3)*(u^3/2) + C, for some constant C.
Reduce the fractions and substitute the u value back in to yield:
(1/9)*[(6x-7)^3/2] + C.
Note: Since the integral is indefinite, you must include the constant C. Teachers will always take off points on tests if you forget this. I hope this helps. Also, you could take the derivative of this to check that you get (6x-7)^(1/2)