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. 
 A: 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.
A: So you used $u = 1 + 7x$. What happens when $x=0$ and $x = 1$? Then you just simply remember that $\frac{d}{dv}(\frac{v^{m+1}}{m+1}) = v^{m}$.
A: \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.
A: This may be useful fou you (Although I haven't written $u$ instead of $7x+1$):

$\int_0^1(1+7x)^{\frac13}dx=\frac17\int_0^1(1+7x)^{\frac13}d(1+7x)=\frac17\times \frac34\int_0^1d(1+7x)^{\frac43}=\frac{3}{28}(1+7x)^{\frac43}\big|_0^1=\frac{3}{28}(8^{\frac43}-1)=\frac{45}{28}.$

