Let $$M=y \; \mbox{and} \; N=x+3x^3y^4$$
We know that our integrating factor will be determined by
$$\mu = e^{\int g(x) \ dx}$$
$g(x)$ will be obtained by assuming that there is a general solution for the differential equation we are trying to solve. $g(x)$ is given by
$$\frac{\frac{\partial M}{\partial y}- \frac{\partial N}{\partial N}}{N}$$
$\Rightarrow \frac{1-(1+9x^2y^4)}{x+3x^2y^4}$
$\Rightarrow - \frac{9xy^4}{1+3x^2y^4}=g(x)$
so far so good, this means our integrating factor will be given by
$\mu = e^{\int \frac{-9xy^4}{1+3x^2y^4}dx}$
$\Rightarrow e^{-\frac{3}{2}ln(1+3x^2y^4)}$
$\Rightarrow \mu = \frac{1}{\sqrt{(1+3x^2y^4)^3}}$
the problem is that my textbook indicates that the integrating factor is $\frac{1}{(xy)^3}$ I can't seem to understand the mistake I'm making any help would be much appreciated
$$y \ dx+(x+3x^{3}y^{4}) \ dy = 0$$ $$ydx+xdy+3x^{3}y^{4}\ dy = 0$$ $$d(xy)+3x^{3}y^{4}\ dy = 0$$ The integrating factor is now obvious: $$\mu (x,y)= \dfrac 1 {(xy)^3}$$ $$\dfrac {d(xy)}{x^3y^3}+3ydy=0$$ Integrate. The integrating factor given is correct. It depends on both $x$ and $y$ not only on $x$. You can find the correct formula for this kind of integrating factor here at section 3 : Formula integrating factor