We will assume that f(x) is a differenciable function on some open interval 'I'.

Therefore, it is countinous.

Therefore, there exists a function F(x) definied on open interval 'I' such that, F'(x)=f(x).

Now we can proceede.

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You are given,

Integral f(x)dx,

You can express it as,

Intergral (1)f(x)dx,

Let u'=x and v=f(x)

Then,

u=1 and v'=f'(x)

By integration by parts,

f(x)-Integral xf'(x)dx