I am reading a textbook about ODE and am confused about the following:
$$\int_0^xf'(t)dt =(t-x)f'(t)\bigg\vert_0^x + \int_0^x (x-t)f''(t)dt $$
Obviously, $$u = f'(t)=\frac{df(t)}{dt} \ \ \text{and} \ \ \frac{dv}{dt}=1$$
So $$\int\frac{dv}{dt}dt=\int1 \Rightarrow v=t+c$$
My question is how to obtain the term $(t-x)$ and $(x-t)$? In the above derivation, I cannot find this term.
Please advise, thanks!
$x$ is fixed, so we can choose the factor $v$ to be equal to $t-x$.