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. 

So you used $u = 1 + 7x$. What happens when $x=0$ and $x = 1$? Then you just simply remember that $\frac{d}{dv}(\frac{v^{m+1}}{m+1}) = v^{m}$.
