Is this integration question correctly answered?

Question

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?

Answer 1

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}

Answer 2

No. You have computed the antiderivative of $3x^2 (x^3 + 2)^7$. Why do you think this informs you about the antiderivative of $(x^3+2)^7$, which is a different function?