The equation of the curve through the point  (1, 2) and whose slope is $\frac{y-1}{x^2+x}$, is

Given $\frac{dy}{dx}=\frac{y-1}{x^2+x}\Rightarrow \frac{dy}{y-1}=\frac{dx}{x(x+1)}$

Integrating we get,

In $(y-1)=2\ln \left(\frac{x}{x+1}\right)+c$

It passes through $(1,2),\text{ so, }c=\log 2$

Required equation is $\ln (y-1)=\ln \left(\frac{2x}{x+1}\right)$

$\Rightarrow (\mathrm{y}-1)(\mathrm{x}+1)-2\mathrm{x}=0$