Question

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

• A. $(y-1)(x+1)+2x=0$
• 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 $(y-1)(x+1)+2x=0$

$m=\frac{y-1}{x(x+1)}=\frac{dy}{dx}$
$\frac{dx}{x(x+1)}=\frac{dy}{y-1}$
$\int dx(\frac{1}{x}-\frac{1}{x+1})=\int \frac{dy}{y-1}$
$lnn-ln(x+1)=ln(y-1)+c$
$ln\mid \left(\frac{x}{x_1}\right)\mid =ln\mid (y-1)\mid +c$
$ln\left(\frac{1}{2}\right)=0+c$
$\frac{x}{x+1}=\frac{y-1}{2}$
$2x+(y-1)(x+1)=0$