Taking constants out of indefinite integrals In the case of definite integrals, the linearity property implies that constants can be taken out of the integrals,
$$\int_{a}^{b} \alpha f(x) d x=\alpha \int_{a}^{b} f(x) d x \tag{1}$$
However, in the case of indefinite integrals, this leads to contradictory results in the case $\alpha=0$, since
$$\int 0\,dx = \int 0 \cdot 1 \,dx = 0 \int 1 \,dx = 0·(x+C) = 0$$
while the derivative of any constant equals $0$, so $\int 0\,dx =C$.
Therefore, can't constants be taken out of indefinite integrals?
 A: That's a good question. Let's see why we say that we can take a multiplicative constant out of an indefinite integral in the first place.
In $\int \alpha f(x) dx$ we are looking for the set of all functions such that the derivative of each one of them is $\alpha f(x)$. If $F(x)$ is such that $F'(x)=f(x)$ then for every $\alpha$ we have that $(\alpha F)'(x)=\alpha f(x)$ by the usual laws of derivatives. This is true even for $\alpha =0$.
However, if $G(x)$ is such that $G'(x)=\alpha f(x)$ then it does not in general follow that $\left(\frac{G(x)}{\alpha}\right)'= f(x)$. It does follow if $\alpha\ne 0$ (again by the usual laws of derivatives), but if $\alpha=0$ the left hand side is undefined. Therefore the set of all antiderivatives of $\alpha f(x)$ and the set $\{\alpha F(x) | F'(x)=f(x)\}$ is not the same set for $\alpha=0$. (More explicitly, the first set is all constants, the second set is just the zero function.)
In other words, you can take out a multiplicative constant out of an indefinite integral, as long as it is non-zero. If it is zero, you have to be careful so that you do not lose all the constant solutions.
