Find the equation of the curve through the point $\left(1,0\right)$ if the slope of the tangent to the curve at any point $\left(x,y\right)$ is $\frac{y-1}{x\left(x+2\right)}.$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhattheslopeoftangenttoacurveat\left(x,y\right)is\\ \frac{dy}{dx}=\frac{y-1}{x^2+x}\Rightarrow \frac{dy}{y-1}=\frac{dx}{x^2+x}\\ Integratingbothsides,wehave\\ {\displaystyle \int \frac{dy}{y-1}={\displaystyle \int \frac{dx}{x^2+x}}}\\ \Rightarrow {\displaystyle \int \frac{dy}{y-1}={\displaystyle \int \frac{dx}{x^2+x+\frac{1}{4}-\frac{1}{4}}}}\left[Makingperfectsquare\right]\\ \Rightarrow {\displaystyle \int \frac{dy}{y-1}={\displaystyle \int \frac{dx}{{\left(x+\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2}}}\\ \Rightarrow log\left|y-1\right|=\frac{1}{2*\frac{1}{2}}log\left|\frac{x+\frac{1}{2}-\frac{1}{2}}{x+\frac{1}{2}+\frac{1}{2}}\right|+logc\\ \Rightarrow log\left|y-1\right|=log\left|\frac{x}{x+1}\right|+logc\\ \Rightarrow log\left|y-1\right|=log\left|c\left(\frac{x}{x+1}\right)\right|\\ \therefore y-1=\frac{cx}{x+1}\Rightarrow \left(y-1\right)\left(x+1\right)=cx\\ Since,thelineispassingthroughthepoint\left(1,0\right),then\\ \left(0-1\right)\left(1+1\right)=c\left(1\right)\Rightarrow c=2\\ Hence,therequiredsolutionis\left(y-1\right)\left(x+1\right)=2x.\end{array}$