Solutions to the differential equation $x(x+1)yy' = xy + 1$ I am having trouble solving the equation
$x(x+1)yy' - xy - 1 = 0$

I will list the steps I followed:
(I'm sure I have made some huge mistake.)



*

*Divide by $x(x+1)$


$yy' - y/(x+1) - 1/x(x+1) = 0$



*Integrate w.r.t dx.


$∫(yy')dx - ∫(y/(x+1))dx - ∫(1/x(x+1))dx = 0$



*Integrating by parts $∫f(x)g'(x)dx = f(x)g(x) - ∫f'(x)g(x)dx$


$f(x) = y$
$g(x) = y$
$∫yy'dx = y.y - ∫y'.y.dx$
$∫yy'dx = y²/2$



*Integrating by parts


$f(x) = y$
$g(x) = \ln(x+1)$
$∫(y/(x+1))dx = y.\ln(x+1) - ∫(dy/dx).\ln(x+1).dx$
$∫(y/(x+1))dx = y.\ln(x+1) - ∫\ln(x+1).dy$
$∫(y/(x+1))dx = y.\ln(x+1) - y.\ln(x+1)$
$∫(y/(x+1))dx = 0$



*Integrating last term


$∫(dx/x(x+1)) = ∫(dx/x) - ∫dx/(x+1)$
$∫(dx/x(x+1)) = \ln(x) - \ln(x+1)$
$∫(dx/x(x+1)) = \ln(x/(x+1))$

$y = √2(\ln(x/x+1))$
$\ln(x/x+1) < 1$ when $0<x<∞$
$y$ becomes imaginary. Judging by that I am sure I have made some huge mistakes, could anyone help me with this. 
 A: $$x(x+1)yy' - xy - 1 = 0$$
Let $y(x)=\frac{1}{z(x)} \quad\to\quad x(x+1)\frac{1}{z} \frac{z'}{z^2} -x\frac{1}{z} -1=0$
$$z'=\frac{1}{x(x+1)} z^3 +\frac{1}{x+1}z^2 $$
This is an Abel's ODE of the first kind. In the present case, as far as I know there is no standard special function available to express the solution on a closed form.
A: Using the substitution $z=x y$, $y'= \dfrac{xz'-z}{x^2}$, gives us
$$(x+1)z\dfrac{xz'-z}{x^2} - z=1$$
or
$$-(x+1)z^2 + x(x+1)zz'- x^2z=x^2$$
Which implies
$$z'=-\dfrac{z}{x}+\dfrac{x}{x+1}+\dfrac{x}{x+1}\dfrac{1}{z}$$
Which is a Chini equation, which tend to be horrible to solve by hand. Wolfram|Alpha also doesn't give a closed form solution.
Maybe someone else sees how to continue, so consider this answer incomplete.
A: $x(x+1)yy'-xy-1=0$
$x(x+1)y\dfrac{dy}{dx}=xy+1$
$y\dfrac{dy}{dx}=\dfrac{y}{x+1}+\dfrac{1}{x(x+1)}$
This belongs to an Abel equation of the second kind.
Let $x=e^t-1$ ,
Then $\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{\dfrac{dy}{dt}}{e^t}=e^{-t}\dfrac{dy}{dt}$
$\therefore e^{-t}y\dfrac{dy}{dt}=e^{-t}y+\dfrac{1}{(e^t-1)e^t}$
$y\dfrac{dy}{dt}-y=\dfrac{1}{e^t-1}$
This belongs to an Abel equation of the second kind in the canonical form.
Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf or in http://www.iaeng.org/IJAM/issues_v43/issue_3/IJAM_43_3_01.pdf
