Differential equation with limit 
Find a solution y=f(x) to the differential equation 
  $$\frac{1}{x^2}y'-\frac{2}{x^3}y+\frac{1}{x^2+x}=0 ,\qquad x>0$$ satisfying
  $\lim_{x\to\infty} \frac{f(x)}{x^2}=1$.

I have $y=x^2(\ln(x+1)-\ln(x)+C)$ so far but can someone please explain the process from beginning to end? Thanks!

$$\lim_{x \to \infty}\frac{x^2(\ln(x+1)-\ln(x)+C)}{x^2}=\lim_{x \to \infty}\ln(x+1)-\ln(x)+C=C+\lim_{x \to \infty}\ln(\frac{x+1}{x})$$
But $\frac{x+1}{x} \to 1$, so $\ln(\frac{x+1}{x}) \to 0$, so:
$$C+\lim_{x \to \infty}\ln(\frac{x+1}{x})=C$$
If you put $C=1$ you have $\lim_{x \to \infty}\frac{y(x)}{x^2}=1$
