Solving integral equation. My answer is wrong. Where do I make a mistake?
$$y(x) = 1+\int_{1}^{x} \frac{y(t)dt}{t(t+1)}dt $$
$$ y'(x) = \frac{d}{dx} \int_{1}^{x} \frac{y(t)dt}{t(t+1)}dt$$
$$ \frac{dy}{dx} = \frac{y(x)}{x(x+1)}$$
$$ \frac{1}{y} = \frac{1}{x(x+1)}dx$$
$$ \int\frac{1}{y} = \int\frac{1}{x(x+1)}dx$$
$$ ln|y| = ln|x|-ln|x+1|+C$$
By y(1) = 1 from the original equation.
$$ C = ln|2| $$
$$ ln|y| = ln|x|-ln|x+1|+ln|2|$$
$$ y = e^{ln|x|-ln|x+1|+ln|2|}$$
$$ y = -x(x+1)ln|2|$$
But the correct answer is $$ y = \frac{2x}{x+1}$$
From $$ y = e^{\ln|x|-\ln|x+1|+\ln|2|}$$ you may write $$ y = e^{\ln|2x|-\ln|x+1|}=e^{\ln\left|\frac{2x}{x+1}\right|}= \frac{2x}{x+1}.$$