partial integration

Question

I have a short question about partial integration. If I want to determine an integral of the form $\int f'gdx$, the formula for partial integration is:

$$\int f'gdx=[fg]-\int fg'dx$$ https://en.wikipedia.org/wiki/Integration_by_parts .

Sometimes it is useful to apply the integration rule twice, for example if $g=x^2$ and then you have to apply partial integration on $\int fg'dx$.

My question is: To calculate $\int f'gdx$, are both possibilities ((1),(2)) allowed:

(1) $\int fg'dx=[Fg']-\int Fg''dx$

(2) $\int fg'dx=[f'g]-\int f'g dx$?

If you do (2) you get back to where you started. – Funktorality Apr 20 '16 at 06:48 — Funktorality, Apr 20 '16 at 06:48
Equation (2) is wrong... How did you get? – Solumilkyu Apr 20 '16 at 07:05 — Solumilkyu, Apr 20 '16 at 07:05

score 1 · Answer 1 · answered Apr 20 '16 at 07:05

1

No. equations 1 & 2 are not valid. IBP is basically the reverse or the re - arrangement of the product rule.

answered Apr 20 '16 at 07:05

user333036

39

score 1 · Answer 2 · answered Apr 20 '16 at 07:27

You can apply the rule as many times as you want/can, because you are just starting an integral afresh every time.

$$\int f'g\,dx=fg-\int fg'\,dx$$ can indeed be followed by

$$\int fg'\,dx=Fg'-\int Fg''\,dx$$

where $F$ is the antiderivative of $f$.

You could also integrate on the other factor, with

$$\int fg'\,dx=fg-\int f'g\,dx$$ but this is of little use.

So a "second order" rule can be written

$$\int f''g\,dx=f'g-fg'+\int fg''\,dx.$$

partial integration

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