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$\int\!{f \left( x \right) \frac{ dg \left( x \right)}{dx}} \ \mathrm{d} x = \begin{vmatrix} u = \frac{f\left( x \right)}{dx}, dv=dg\left( x \right) dx \\ du=\frac{df\left( x \right)}{dx} dx, v=g\left( x \right) \end{vmatrix} = \frac {f \left( x \right) g \left( x \right)}{ dx } - \int\!{g\left( x \right) \frac {df\left( x \right)}{dx}} \ \mathrm{d} x$

I'm a little bit confused by this $dx$ denominator.

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up vote 1 down vote accepted

Integration by parts is just the product rule for derivatives, rearranged. Start with $$(fg)'=fg'+f'g$$ make it $$fg'=(fg)'-f'g$$ and integrate: $$\int f {dg\over dx}\,dx=fg-\int g{df\over dx}\,dx$$

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Oops! Need to sleep more. :D It's just a definition of this rule... Well, actually even with my plain stupid substitution I can end up with the same result. The mistake is in du and v. du = df(x)/dx and v = g(x)dx. – Wildcat Feb 6 '12 at 10:47

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