Integration by parts- using a u and v that are not inside of the original integral? For instance, if I want to integrate some function $\frac{df(x)}{dx}=f'(x)$
$$\int_a^b f'(x)\,dx$$
And I use integration by parts, is it acceptable to set $u=x$ and $dv = f'(x)$ even though $x$ isn't in my integral?
 A: By definition $f^{\prime}(x)$ has $x$ in it (if it's a constant then integration by parts is not needed). The point of integrating by parts is to take the part of $f^{\prime}(x)$ which is going to make your integrand simpler when it has been differentiated, and set that as $u$.
For example, if your integrand is $x^2 \sin(x)$ you could set $u=x^2$ so that $du = 2x \, dx$. And set $dv = \sin(x) \, dx$ so that $v = -\cos(x)$. 
Integration by parts then says that we take the original integral is equal to $\displaystyle\left.uv\vphantom{\frac11}\right|_a^b - \int_a^bv\,du$
You can see that the new integrand ($-2x\,\cos(x)\,dx$) is one step closer to being a simply integrable function. One more application of integration by parts and you're there. 
Hopefully this illustrates why picking a parameter that isn't in your function to begin with is not so useful. 
A: You'd need $u=1$ and $dv=f'(x)\,dx$, so that $du=0\,dx$ and $v=f(x)$.  Then you have
$$
\int f'(x)\,dx = \int u\,dv=uv-\int v\,du = 1\cdot v - \int f(x)\cdot 0\, dx = v +C = f(x) + C.
$$
But integration by parts is overkill for this problem.
