Q:

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 $$z=2\int \frac{dt}{\left(1+t^2\right)^2}$$

