Number of solutions to a differential equation The number of solutions of $$\frac{dy}{dx} = \frac{y+1}{x-1}$$ when $y(1) = 2$ is

Options are a) none   b) one   c) two   d) infinite

Now this can be solved via variable separable method which gives me
$$\frac{dy}{y+1} = \frac{dx}{x-1}$$
And on integrating both sides I got,
$\log (y + 1) = \log(x - 1) + \log c$
which gave me
$y + 1 = c(x - 1)$
On putting the given values as $x = 1$ and $y = 2$ gives
$3 = 0$ or $c = \frac{3}{0}$ which means no solution.

I just want to ask if my solution is correct because the book gave the answer as one solution. I checked few other places and they for some reason are solving by integrating as
$\log (y + 1) = \log(x - 1) - \log c$
which of course would give them the desired one solution as $x = 1$. 

I am confused.
The problem is ill-posed since $x=1$ is a singular point. But even if you consider the given point to be a limit, you can see from the general solution that

$$ \lim_{x\to 1} y(x) = \lim_{x\to1} \big[c(x-1)-1 \big] = -1 $$

Clearly, there are no such functions where $\lim_{x\to1} y(x) = 2$
If you consider $x(y)$ not $y(x)$ you'll get $x = 1 + (1 + y)\cdot c$, and therefore substituting $y=2$ you'll get $c=0$ if you want $x=1$. The solutions is therefore $x(y)=1$.

Edit: also mind that $\int dx/x=\ln|x|$.