4xydx +(x^2+1)dy = 0
$\displaystyle 4xydx+(x^{2}+1)dy=0$
$\displaystyle (x^{2}+1)dy=-4xydx$
$\displaystyle \frac{dy}{y}=\frac{-4x}{x^{2}+1}dx$
$\displaystyle \int\\\frac{dy}{y}=\int\\\frac{-4x}{x^{2}+1}dx$
The left hand side becomes $\displaystyle ln(y)$.
For the right hand side, let $\displaystyle u=x^{2}+1$, then
$\displaystyle du=2xdx\Rightarrow\\\frac{du}{2}=xdx$
So, $\displaystyle \int\\\frac{-4x}{x^{2}+1}=\frac{-4}{2}\int\\\frac{du}{u}$
$\displaystyle =-2ln(u)+C=-2ln(x^{2}+1)+C$
We are left with,
$\displaystyle ln(y)=-2ln(x^{2}+1)+C$. Solving for y we have,
$\displaystyle y=\frac{e^{C}}{(x^{2}+1)^{2}}$.
you can fill in the gaps.
$\displaystyle e^{-2ln(x^{2}+1)+C}=e^{c}e^{-2ln(x^{2}+1)}=
e^{c}(x^{2}+1)^{-2}=\frac{e^{c}}{(x^{2}+1)^{2}}$
Remember your log properties. $\displaystyle log(x^{c})=clog(x)$. So, in this case, e cancled out the natural log and the -2 out front jumped back up to its exponent position.