In case you're wondering, this part is taken from the answer key of my textbook, and was in the midst of an integrating by substitution solution, hence the u, (I just didnt write the whole thing out)
$$\int\frac{u^2-u}{u+1}du = \int(u-2)+\frac{2}{1+u}du$$
So this is the part which I don't seem to grasp on; how does the left side of the equation turn into the right side? I seem to know how they did it, it looks to me they trialed and error with the cross factorisation $(u^2-u)$ and then $-2+2$ such that it factorise nicely with it being $(u+1)(u-2)$.
Was that how they did it? Or is there a much cleaner way of doing that?
Alternatively, if I dont follow their answer key, the only method I can think of is a double substitution where $v=u+1$. Are there other ways?
\begin{align} u^2 - u &= u(u+1)-u-u \\ &= u(u+1) -2u \\ &= u(u+1) -2(u+1-1) \\ &= u(u+1) -2(u+1)+2 \\ &= (u-2)(u+1)+2 \end{align}
Now, divide both sides by $u+1$.