Why does integration by parts work?

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Answer 1 · 2018-04-04T13:19:36

Because of the product rule for differentiation.

Recall that the integral of a function is the function (family) that has that derivative.

$int f(x) dx = F(x) +C$ if and only if $F'(x) = f(x)$ .

That is to say

$int F'(x) dx = F(x) +C$ .

We know from our study of derivatives that

$d/dx(f(x)g(x)) = f(x) g'(x) + g(x) f'(x)$ .

Written in terms of differentials, we have:

$d(uv) = u dv + v du$ .

So, with $uv$ in the role of $F(x)$ above, we have

$int d(uv) = int (udv+vdu)$ .

So,

$uv = intudv + intvdu$ .

And

$intudv + intvdu = uv$ .

Consequently,

$intudv = uv-intvdu$ .

Written using prime notation, we have

$d/dx(f(x)g(x)) = f(x) g'(x) + g(x) f'(x)$ .

$f(x)g(x) = int f(x) g'(x)dx + intg(x) f'(x)dx$

$int f(x) g'(x)dx + intg(x) f'(x)dx = f(x)g(x)$

$int f(x) g'(x)dx = f(x)g(x)- intg(x) f'(x)dx$