How to solve $\int^1_0 (1+7x)^{1/3}dx$? I worked through $\int^1_0 (1+7x)^{1/3}dx$ and I got $\frac{3}{4}+C$ for the answer. However, I forgot about the exponent when I found the difference of the sides of the integral.
I am retrying it, but I've realized I don't know how to find a number to the power of $\frac{4}{3}$. Also, when I went over this with Symbolab, once u-substitution had been applied, the integral changed to $\int^8_1$ for some reason. I'm sure it's the key to solving this, but I have no idea why that's even allowed.
My textbook and Symbolab both say the answer is $\frac{45}{28}$.
Here are the steps I took. Please let me know what I got wrong.

$\int^1_0 (1+7x)^{1/3}dx$

Let $u=1+7x$
Then $du=7dx$ and $dx=\frac{1}{7}du$
so $\int^1_0u^{1/3}\frac{1}{7}du=\frac{1}{7}\int^1_0u^{1/3}$
$$\frac{1}{7}\int^1_0u^{1/3}$$
$$=\frac{1}{7}[\frac{u^{4/3}}{4/3}|^1_0]$$ $$=\frac{1}{7}[\frac{3u^{4/3}}{4}|^1_0]$$
$$=\frac{1}{7}[\frac{3(1+7(1))^{4/3}}{4}-\frac{3(1+7(0))^{4/3}}{4}]$$
$$=\frac{1}{7}[\frac{3(1+7)^{4/3}}{4}-\frac{3(1+0)^{4/3}}{4}]$$
This is as far as I can get. 

You got $\dfrac{1}{7}\left[\dfrac{3(1+7)^{4/3}}{4}-\dfrac{3(1+0)^{4/3}}{4}\right].$
Note that $(1+7)^{4/3}=8^{4/3}=(8^{1/3})^4=2^4=16$ and $(1+0)^{4/3}=1^{4/3}=1,$
so you actually got the correct answer:  $\dfrac17\left[\dfrac{3\times16}4-\dfrac{3\times1}4\right]=\dfrac17\dfrac{45}4=\dfrac{45}{28}.$
