Given :The slope of the tangent to the curve at any point is $\large\frac{y-1}{x^2+x}$

The slope of the tangent to the curve is $\large\frac{dy}{dx}$

Therefore $ \large\frac{dy}{dx}=\frac{y-1}{x^2+x}$

Seperating the variables we get

$ \large\frac{dy}{y-1}=\frac{dx}{x(x+1)}$

Now $\large\frac{1}{x(x+1)}$ can be resolved into paritial fractions as $\large\frac{A}{x}+\frac{B}{x+1}$

Hence $1=A(x+1)+B(x)$

Now let us equate the coefficient of like terms

First let us equate the 'x' term

$0=A+B$-----(1)

Equating the constant term

$1=+A\qquad B=-1$

Hence $\large\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}$

Therefore $\large \frac{dy}{y-1}$$=+\large\frac{dx}{x}-\frac{dx}{x+1}$

Integrating on both sides we get,

$\int \large \frac{dy}{y-1}$$=+\int \large\frac{dx}{x}-\int \frac{dx}{x+1}$

=>$ \log (y-1)=+\log x-\log (x+1)+\log c$

=>$\log (y-1)=\log \large\frac{c(x)}{(x+1)}$

=>$(y-1)=\Large\frac{cx}{x+1}$

=>$(y-1)(x+1)=cx$

To evaluate the value of c,let us now substitute the value of x and y with the given point (1,0)

$(0-1)(1+1)=c(1)$

=>$c=-2$

$(y-1)(x+1)=-2x$

$(y-1)(x+1)+2x=0$ is the required solution.