If $$y(x-y)^2 = x.$$
Then $$\int \frac{1}{x-3y} \, dx =? $$
How do I go about doing This?
Just if you want to know the options are in the form of log{(x-y)² + 1}
[This may not be the answer as this is just one of the options, Its just to get an idea]
Hints will do.