Correct way to integrate $\int x(x^2-16)dx$

Question

Evaluate: $$\int x(x^2-16)dx$$

I have noticed that this integral can be solved using two different methods, but I am not sure which one is the correct one.

Way 1: Using $u$-subtitution

Let $u=x^2-16, du = 2xdx$
Then, we have $$\int x(x^2-16)dx$$ $$= \frac{1}{2}\int udu$$ $$= \frac{1}{2}(\frac{1}{2}u^2)+C$$ $$= \frac{1}{4}(x^2-16)^2+C$$

Way 2:

$$\int x(x^2-16)dx$$ $$= \int(x^3-16x)dx$$ $$= \frac{1}{4}x^4-\frac{16}{2}x^2+C$$ $$= \frac{1}{4}x^4-8x^2+C$$

Answer 1

Both methods are correct. Notice that $$\frac14(x^2-16)^2=\frac14(x^4-32x^2+256)=\frac14x^4-8x^2+64.$$

The $C$ in your first way is simply $64$ less than the $C$ in your second way.

Answer 2

Alternatively, use hyperbolic trigonometric substitution.

Let $x = 4\cosh t\Rightarrow x^2 - 16 = 16\sinh^2 t$ and $\frac{dx}{dt} = 4\sinh t$.

The integral becomes

$$\begin{array} {r c l } \displaystyle \int 4\cosh t \cdot (16\sinh^2 t)(4\sinh t) \, dt &=& 4^4 \displaystyle \int (\cosh t)(\sinh t)^3 \, dt = 4^3 \sinh^4 t + C \\ &=& 64(\cosh^2 + 1)^2 + C \\ &=& 64\left(\left ( \frac x4 \right)^2 + 1\right)^2 + C \end{array} $$