Find the general solution to differential equation $x(x+1)^2(y'-\sqrt x)=(3x^2+4x+1)y$ Equation can be transformed to linear differential equation:
$$LHS=x^3y'+2x^2y'+xy'-x^{7/2}-2x^{5/2}-x^{3/2}$$
$$RHS=3x^2y+4xy+y$$
$$\Rightarrow y'(x^3+2x^2+x)+y(-3x^2-4x-1)=x^{7/2}+2x^{5/2}+x^{3/2}$$
After dividing by $(x^3+2x^2+x),x\neq 0\land x\neq -1$
$$\Rightarrow y'+\frac{-3x^2-4x-1}{x^3+2x^2+x}y=\frac{x^{7/2}+2x^{5/2}+x^{3/2}}{x^3+2x^2+x}$$
The general solution is
$$y=e^{-\int \frac{-3x^2-4x-1}{x^3+2x^2+x}\mathrm dx}\left(c+\int \frac{x^{7/2}+2x^{5/2}+x^{3/2}}{x^3+2x^2+x}{e^{\int \frac{-3x^2-4x-1}{x^3+2x^2+x}\mathrm dx}}\mathrm dx\right)$$
Integral $\int \frac{-3x^2-4x-1}{x^3+2x^2+x}\mathrm dx$ can be found using partial fractions, $$\int \frac{-3x^2-4x-1}{x^3+2x^2+x}\mathrm dx=\ln(|x|(x+1)^2)+c$$
How to evaluate integral 
$$\int \frac{x^{7/2}+2x^{5/2}+x^{3/2}}{x^3+2x^2+x}{e^{\int \frac{-3x^2-4x-1}{x^3+2x^2+x}\mathrm dx}}\mathrm dx=\int \frac{x^{7/2}+2x^{5/2}+x^{3/2}}{x^3+2x^2+x}e^{\ln(|x|(x+1)^2)}\mathrm dx?$$
Is there an easier method than transforming to linear equation?
 A: What seems to be interesting is to define $$y=z\, x\,(1+x)^2$$ which makes the differential equation to be $$x (x+1)^2 \,z'-\sqrt{x}=0$$ which is separable and "quite" simple to integrate. $$z=\int \frac{\sqrt{x}}{x (x+1)^2}\,dx$$ Make $x=t^2$ to get $$z=2\int \frac{dt}{\left(1+t^2\right)^2}$$
A: Starting from the equation in the title, we have
$$x(x+1)^2y'-(3x^2+4x+1)y=x(x+1)^2\sqrt x$$
It just so happens we have
$$[x(x+1)^2]'=3x^2+4x+1$$
This suggests use of the quotient rule.
$$\dfrac{x(x+1)^2y'-(3x^2+4x+1)y}{x^2(x+1)^4}=\left[\dfrac{y}{x(x+1)^2}\right]'=\dfrac{\sqrt x}{x(x+1)^2}$$
One messy integral left and we should have the solution.  Next step would likely be partial fractions, although substitution may be possible.
$$x=\tan^2\theta,dx=2\tan\theta\sec^2\theta d\theta$$
$$\int\dfrac{\sqrt xdx}{x(x+1)^2}=\int\dfrac{2\tan^2\theta\sec^2\theta d\theta}{\tan^2\theta\sec^4\theta}=\int\cos^2\theta d\theta=\int(\frac12+\frac12\cos2\theta)d\theta=$$
$$\frac12\theta+\frac14\sin2\theta$$
That just leaves a messy backsubstitution, quite possibly using the power reduction formula for tangent.
