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. 

\begin{align}
\int\limits_0^1 (1+7x)^{1/3} dx 
&= \int\limits_{u(0)}^{u(1)} u^{1/3}\frac{1}{7} du 
& & \mbox{substituting } u = 1+7x \Rightarrow du = 7 dx \\
&= \left[ \frac{3}{4}\frac{1}{7} u^{4/3} \right]_1^8
& & \mbox{integration rule for } x^r \mapsto ({\small r+1})^{-1} x^{r+1} \\
&= \frac{3}{28}(16-1) \\
&=\frac{45}{28}
\end{align}
WolframAlpha link here.
