Partial integration of $\sin x\log(y-1)$ w.r.t. $x$

Question

If I have the function $\sin x\log(y-1)$ and I want to partially integrate it w.r.t. $x$ then what happens to $\log $? Would the solution be: $-\cos x \log(y-1)$ and how? Isn't $\log(y-1)$ a function or a constant?

Answer 1

Yes, if you are concerned with the function $(x,y) \mapsto \sin x \log |y-1|$.

Let $y \neq 1$ and let $f: x \mapsto \sin x \log |y-1|$ on $\mathbb{R}$. Then $$ \int_{x} \sin x \log |y-1| = -\log|y-1|\int_{x} D\cos x = -\cos x \log|y-1| + \text{some constant}. $$

But if you are concerned with the function $x \mapsto (x, f(x)) =: (x,y) \mapsto \sin x \log |y-1|$, then of course you cannot simply treat $y$ as a constant...