Note: I use $\cdot$ to signify multiplication.
We could either have:
$$\int \frac{dH}{kS-kH} = \frac1k \int \frac{dH}{S-H} = \frac1k \cdot -\ln|S-H|+C$$
Or we could use $u$-substitution instead:
$$u=kS-kH \Rightarrow du =-k\cdot dH \Rightarrow \frac{du}{-k} =dH $$
$$\int \frac{dH}{kS-kH} = \int \frac1u \frac{du}{-k}= \frac1{-k}\int \frac1u du = \frac1{-k} \ln|u|+C = \frac1{-k} \ln|kS-kH|+C$$
As far as I can tell, $$ \frac1k \cdot -\ln|S-H|+C \neq \frac1{-k} \ln|kS-kH|+C$$
because $$\ln|kS-kH| = \ln|k|+ \ln|S-H| \neq \ln|S-H|$$
so I have a contradiction. (Unless $k=\pm e$, which isn't guaranteed.)
What am I doing wrong?
There is nothing going wrong in either of the cases.
The value constant of integration in the two cases is different, so it will not make any contradiction.
C1=C2-ln (k)/k
See that ln (k) is a constant.