Question

The equation of the curve through the point $(1,0)$ and whose slope is $\frac{y-1}{x^2+x}$ is

• A. $\left|\frac{(y-1)(x+1)}{x}\right|=2$
• B. $2x(y-1)+x+1=0$
• C. $x(y-1)(x+1)+2=0$
• D. None of these

Answer 1

The correct option is A $\left|\frac{(y-1)(x+1)}{x}\right|=2$

$\frac{dy}{dx}=\frac{y-1}{x^2+x}$
$\Rightarrow \frac{dy}{y-1}=\frac{dx}{x^2+x}$
$\Rightarrow \int \frac{1}{y-1}dy=\int (\frac{1}{x}-\frac{1}{x+1})dx$
$\Rightarrow \ln |y-1|=\ln |x|-\ln |x+1|+\ln |c|$
$\Rightarrow \left|\frac{(y-1)(x+1)}{x}\right|=|c|$
Putting $x=1,y=0,$ we get $c=\pm 2$
The equation is$\left|\frac{(y-1)(x+1)}{x}\right|=2$