Integration with complex constant

Yes, this is true; but why this is true is bit more delicate question.

To keep it simple: one can define integrals of complex-valued functions along curves in the complex plane. The resulting integral-notion will be $\Bbb C$-linear (as one might expect) instead of only $\Bbb R$-linear (as the usual, say, Riemann-integral is). However, this more general integral notion still includes the usual one as special case and hence the more general linearity follows at once.

(Note: Usually, complex integrals are defined as line integrals and not as general expressions involving an integral sign. But, intuitevely, I do not think this should matter here.)

I agree with the answer given by mrtaurho.

I just want to add that $\mathbb{C}$-linearity still holds if you are interested in finding an antiderivative of a given complex valued function, which seems to be your case to me.

This is quite obvious if you think about it: you can picture $\mathbb{C}$ simply as $\mathbb{R}^2$ endowed with a product operation that gives it a field structure. Then, functions $\mathbb{C}\rightarrow\mathbb{C}$ are just functions $\mathbb{R}^2\rightarrow\mathbb{R}^2$, and you have just traced back your problem to real numbers.

However, the problem of finding an antiderivative in several variables is not as straightforward as in one variable, and it is better understood in terms of closed and exact differential forms.

But to keep it simple and short: yes, complex constants behave in the exact same way.