Is this integration question correctly answered? So we have:
$$y = (x^3 + 2)^7$$
And we want to find:
$$I = \int y \space dx \\$$
From the chain rule, we know:
$${d\over dx} g \Big( f(x) \Big) = g' \Big( f(x) \Big) \cdot f'(x)$$
Hence:
$$\int g' \Big( f(x) \Big) \cdot f'(x) \space dx =  g \Big( f(x) \Big) \\ $$
Let's call this identity $(1)$ so we don't have to rewrite it.
So I looked at  $\space (x^3 + 2)^8 $. Which differentiates to give:
$${24x^2} (x^3 + 2)^7$$
So if we integrate this we get:
$$8 \int {3x^2} (x^3 + 2)^7 \space dx= (x^3 + 2)^8 $$
(Note I took $8$ out the integral).
Now we divide by $8$ and get:
$$\int {3x^2} (x^3 + 2)^7 \space dx= {(x^3 + 2)^8 \over 8}$$
The integral on the left is in the same format as the identity $(1)$.
Does this mean that:
$$I = {(x^3 + 2)^8 \over 8} + C$$
Or have I gone wrong somewhere?

You can apply binomial theorem to integrate $(x^3+2)^7$ directly:
\begin{align}
\int (x^3+2)^7 dx &= \int \left(\sum_{k=0}^7\binom{7}{k} 2^{7-k}x^{3k}\right) dx\\
&=\sum_{k=0}^7 \binom{7}{k}\frac{2^{7-k}}{3k+1}x^{3k+1}+C
\end{align}
