normal distribution derivation In this derivation:
http://www.sonoma.edu/users/w/wilsonst/Papers/Normal/default.html
how do these equal?
$$ -k\int (x-\mu) dx = -\frac{k}{2} (x-\mu)^2$$ 
Isn't this the case?
$$ -k\int (x-\mu) dx = -\frac{kx^2}{2} + k\mu x$$ 

The answer given comes from one antiderivative of $x-\mu$.  Your answer comes from another antiderivative of $x-\mu$. The two differ by a constant, so both of them are correct antiderivatives. Neither of them is the general antiderivative of $x-\mu$.  
The general antiderivative of $x-\mu$ can be written as $\frac{1}{2}(x-\mu)^2+C$ or as $\frac{x^2}{2}-\mu x+C$, where in each case the $C$ is an arbitrary constant of integration. Your source presumably found the first form more convenient.  
