0

Pardon the fact that I may be butchering the proper name of the general rule for the below indefinite integration in the title (and please feel free to edit the heading):

$$\int((f(x))(g(x)))\>dx\>=\>g(x)*\int(f(x)\>dx)-\int[\int((f(x)\>dx)\>*(\frac{d}{dx}g(x))]\>dx$$

I've been struggling conceptualizing the rule and would love if someone could logically and pre-algebraically prove it.

sardinsky
  • 133

1 Answers1

0

Imagine the product rule for derivatives $$(uv)' = u' v + u v'.$$ If we integrate and rearrange we obtain the rule for integration by parts. $$ \int u' v \, dx = uv - \int uv' \, dx. $$

Gregory
  • 3,641
  • Do you mean the integral of $u\prime$ becomes u. And can you further explain how the second term right of the equals sign in the most bottom expression is equivalent to the second term in the rule I have posted above. i.e. how does $\int((u)(v\prime))dx$ equate to $\int[\int((f(x)>dx)>*(\frac{d}{dx}g(x))]>dx$ – sardinsky Oct 25 '17 at 23:59
  • Define $u' = f$ and $v = g$ and write it out for yourself, they should be the same. – Gregory Oct 26 '17 at 15:44