Integrate $(1+7x)^{1/3}$ from $0$ to $1$ 
Integrate $(1+7x)^{1/3}$ from $0$ to $1$

So using substitution, I'm able to get to $$\frac17  \frac34 u ^{4/3}$$ Pretty sure that part is right, but I'm getting stuck after that. 

It looks like you did the integration correctly.  Now you just have to evaluate your answer at the upper and lower limits.
The problem is that the upper and lower limits 1 and 0 are for the variable $x$.  You did a variable change so that your integral became in terms of $u$.  This means you have to evaluate your answer at the upper and lower limits of the variable $u$.
But how do you find the upper and lower limits of the variable $u$?  Well, to solve the problem you had set $u = 7x + 1$.  And the lower limit for the variable $x$ was $0$.  Plugging in this value of $x$ into your equation $u = 7x + 1$ will give you the new lower limit of $u$.
Similarly, to find the upper limit of $u$, we know the upper limit of $x$ was $1$, so plugging this into $u = 7x + 1$ will give you the new upper limit.
So, if you did the work correctly, your new upper and lower limits should be 8 and 1.  So you must evaluate your answer at $u = 8$, and then subtract what you get when you evaluate it at $u = 1$, and that is your final answer.
