I have a separable equation $\frac{dy}{dx}$=$\frac{27}{y^{1/3}+81x^2y^{1/3}}$
I separated both sides by multiplying by dx and factoring out the y^{1/3} and multiplying, as well. Then I integrated with respect to x and y.
At first I didn't notice that integrating would give arctan so instead I moved 27 outside of the integral and integrated the inside as ln(1+81x^2).
Is using ln also equivalent to using arctan?
No;
$$\int\frac{dx}{1+81x^2}=\frac19\arctan(9x)+C$$
By using $$\int\frac{dy}{1+y^2}=\frac{1}{y'}\arctan(y)+C$$
You can't use $\ln$ here because the denominator of the integrand is not linear. Also, the $27$ just means you need to multiply the above by $27$ and will never change the antiderivative by anything more than a multiple of it. Essentially: $$\int kf(x) \ dx = k\int f(x) dx$$ for a constant $k$ is worth remembering
For reference; $$(\ln(1+81x^2))'=\frac{(1+81x^2)'}{1+81x^2}=\frac{162x}{1+81x^2}$$ and you can see why you can't use $\ln$ when the denominator isn't linear.
