I am trying to find the indefinite integral wrt $x$ of $\delta(ct-x)$; so I let $u=ct-x$, then $du/dx=-1$ and $dx=-du$ so the integral becomes
$$-\int \delta(u)du = -\theta(u) = -\theta(ct-x) $$
but Wolfram Alpha Pro gives $\theta(x-ct)$ as the result. Note the variables are swapped.
Here $\theta$ is the Heaviside step, $\delta$ is the Dirac delta distribution, $x$ and $t$ are variables, and $c$ is a constant $>0$.
I checked my answer as follows
$$d/dx (-1)\theta(ct-x)=-\delta(ct-x)(-1)=\delta(ct-x) $$
so it checks out (I think).
Checking Wolfram's answer gives
$$d/dx (\theta(x-ct))=\delta(x-ct)= \delta(ct-x) $$
so it also checks out.
MY QUESTION IS: who is right, Wolfram Alpha or me? Or both and am I missing an identity?
The two answers differ by a constant which is why the derivatives are the same. Note that $\theta(x) = -\theta(-x)+1$. So the two expressions you and wolfram got are not equal, they differ by 1, but since you are taking an indefinite integral the difference is in the $+C$.