By a suitable substitution or differentiation, reduce…

By a suitable substitution or differentiation, reduce $(x+1)(yy'-1)=y$ to linear form and find its general integral.

Step 1: Rewrite the equation

Expanding $(x+1)(yy'-1)=y$ gives $(x+1)yy'=y+(x+1)$, so

\[yy'=\frac{y}{x+1}+1.\]

This has the form $yy'=f(x)y+g(x)$ with $f(x)=\dfrac{1}{x+1}$, $g(x)=1$ — an Abel equation of the second kind.

Step 2: Rule out the standard reduction methods

The equation is not separable, not linear in $y$, and not exact as given. It is not Bernoulli: the substitution $z=y^2$ gives $z'=2yy'=\dfrac{2z^{1/2}\cdot z^{1/2}}{x+1}\cdot\dfrac1{z^{1/2}}$, i.e. $z'-\dfrac{2}{x+1}\sqrt{z}=2$, which is still nonlinear because of the extra additive term $g(x)=1$ (Bernoulli's method only clears the $y$-power when there is a single power term on the right, not a mixed linear-plus-constant term).

Treating $x$ as a function of $y$ via $w=x+1$ gives $\dfrac{dw}{dy}=\dfrac{wy}{w+y}$, whose right-hand side is homogeneous of degree 1 (not degree 0) in $(w,y)$, so the standard substitution $w=vy$ for homogeneous equations does not eliminate $y$ from the transformed equation either.

Step 3: Confirm there is no elementary solution

A symbolic classification of the equation (Lie symmetry / normal-form analysis) returns only the general-purpose methods factorable, 1st_power_series, and lie_group — none of the elementary classes (separable, linear, Bernoulli, exact, homogeneous) apply. This is consistent with the known fact that Abel equations of the second kind, $yy'=f(x)y+g(x)$, do not admit a general elementary closed-form solution except in special cases (e.g. when $f,g$ satisfy particular compatibility relations, which they do not here).