Is it normal that u-substitution leaves behind a constant besides C?

Question

Hello in the following problem $\int{(x^2+2)(x^3+6x+1)^5}$, using u-substitution we get the following: $\frac{(x^3+6x+1)^6}{18} + C$, but when we simplify this in the constant end we get $1/18$ alongside $C$, however if we do the integral of the question but fully simplified($x^{17}+.....+2$), we do not get the $1/18$, in fact we don't get any constant. Why are these integrals different? Is there some limitations of u-sub I should know about?

Answer 1

in fact we don't get any constant.

How is that? You should still have the constant of integration and in fact because of this, the difference you think is there, actually isn't...

Why are these integrals different?

When we write $$\int \left( x+1 \right) \,\mbox{d}x = \color{green}{\frac{x^2}{2}+x}+\color{blue}{C} \quad\quad \left(C \in \mathbb{R}\right)$$ the " $+C$ " makes sure we didn't write an antiderivative of $x+1$, but (all) the antiderivatives.

Now compare with using a (simple) $u$-substitution ($u=x+1$) for the same problem: $$\int \left( x+1 \right) \,\mbox{d}x =\int u \,\mbox{d}u = \frac{u^2}{2}+C= \frac{\left( x+1 \right)^2}{2}+C= \color{green}{\frac{x^2}{2}+x}+\color{red}{\frac{1}{2}+C} \quad\quad \left(C \in \mathbb{R}\right)$$

It may not look the same, but we get the exact same (set of) antiderivatives.

To get one specific antiderivative, the values of the "blue $C$" and "red $C$" will differ (by $1/2$).

Answer 2

The answers in both the cases are same. As you can always club the constant of integration with the constant value.
$\frac{1}{18} + C$ can be called a new constant $C_1$.