Find the solution to the differential equation Assume $x\gt 0$ and let 
$$x(x+1)\frac{du}{dx} = u^2,$$
$$u(1) = 4.$$
I started off by doing some algebra to get:
$$\frac{1}{u^2}du = \frac{1}{x^2+x}dx.$$
I then took the partial fraction of the right side of the equation:
$$\frac{1}{u^2}du = \left(\frac{1}{x}-\frac{1}{x+1}\right).$$
I then took the integral of both sides:
$$-\frac{1}{u} = \log{x}-\log{(x+1)}+C.$$
From here I don't know what to do because we are solving for $u(x)$ and I'm not sure how to get that from $-\frac{1}{u}$.

Solving for $\frac{du}{dx}$ we have
\begin{equation*}
\frac{du}{dx}=\frac{u^2}{x(x+1)}\\
\Rightarrow \frac{du}{dx}=\frac{u^2}{x^2+x}\\
\frac{\frac{du}{dx}}{u^2}=\frac{1}{x^2+x}.
\end{equation*}
Integrate both sides & evaluate the integrals:
\begin{equation*}
-\frac{1}{u}=\log(x)-\log(x+1)+C_1\\
\Rightarrow u=-\frac{1}{\log(x)-\log(x+1)+C_1}.
\end{equation*}
Now apply the initial condition:
\begin{equation*}
-\frac{1}{C_1-\log(2)}=4\Rightarrow C_1=\frac{1}{4}(-1+4\log(2)).
\end{equation*}
This gives the result. 
