Differentiation behaves well with respect to sums and products with numbers

$$(aF(x)+bG(x))'=aF'(x)+bG'(x)$$

(we say it is *linear*). So integration behaves the same way: if

$$\int f(x)\,\mathrm{d}x=F(x)+k$$

$$\int g(x)\,\mathrm{d}x=G(x)+k$$

then

$$\int (af(x)+bg(x))\,\mathrm{d}x=aF(x)+bG(x)+k$$

($k$ simply denotes the constant degree of freedom, so sums and products don't apply on it)