Integrating $\int{\frac{x}{x+1}}\,dx$ None of the methods I know can solve this easily. I tried integrating by parts, but it essentially leads you to the equation $\int{\frac{x}{x+1}}\,dx=\int{\frac{x}{x+1}}\,dx$, which isn't very helpful.
Using Wolfram Alpha, it seems like you need to rewrite the integrand:
$$\frac{x}{x+1}=1-\frac{1}{x+1}$$
That makes it easy to integrate. Is this some method that others know about? I never came across a problem like this before, so it's interesting for me. In what cases should one rewrite the integrand like this in order to make integration easier? I'm asking this so that I can be prepared for the next time a question like this stumps me.

This is a standard technique: $$\frac{x}{x+1} = \frac{x+(1-1)}{x+1}=\frac{x+1}{x+1}-\frac{1}{x+1}$$
Notice all I did was to add $0$.
