How to find integrating factor for this differential equation Solve this differential equation $dx/dt=x^2t^3+xt$
This is not exact.I don't find any way to solve this by integrating factor
 A: $$\frac{dx}{dt}=x^2t^3+xt$$
$$(x^2t^3+xt)dt-dx=0$$
Solving thanks to the integrating factor method.
Let $\mu(x,t)$ an integrating factor.
$$(x^2t^3+xt)\mu dt-\mu dx=0$$
The condition to be an exact differential is :
$$\frac{\partial}{\partial x}\left((x^2t^3+xt)\mu  \right) = \frac{\partial}{\partial t}(-\mu)$$
$$\frac{\partial \mu}{\partial t}+(x^2t^3+xt)\frac{\partial \mu}{\partial x}+(2xt^3+t)\mu=0$$
In the most simple cases $\mu(x,t)$ is on the form $f(x)$ or $g(t)$ or $h(xt)$. But in the present case, trying those forms of $\mu(x,y)$ fails. So, we try a more general form, for example 
$\mu(x,t)=f(x)g(t)$
$$f(x)\frac{dg}{dt}+(x^2t^3+xt)g(t)\frac{df}{dx}+(2xt^3+t)f(x)g(t)=0$$
$$\frac{1}{tg(t)}\frac{dg}{dt}+(xt^2+1)\frac{x}{f(x)}\frac{df}{dx}+(2xt^2+1)=0$$
The separation of variables becomes possible if 
$$xt^2\frac{x}{f(x)}\frac{df}{dx}+2xt^2=0\quad\implies\quad \frac{x}{f(x)}\frac{df}{dx}=-2\quad\implies\quad f(x)=\frac{1}{x^2}$$
Then
$$\frac{1}{tg(t)}\frac{dg}{dt}+\frac{x}{f(x)}\frac{df}{dx}+1=0 \quad;\quad \frac{1}{tg(t)}\frac{dg}{dt}-2+1=0$$
$$\frac{1}{g(t)}\frac{dg}{dt}=t\quad\implies\quad g(t)=e^{t^2/2}$$
An integrator factor is :
$$\mu(x,t)=\frac{e^{t^2/2}}{x^2}$$
From this it is easy to solve the ODE. The result is :
$$e^{t^2/2}\left(\frac{1}{x}+t^2-2 \right)=c$$
$$x(t)=\frac{-1}{2-t^2+c\:e^{-t^2/2}}$$
