[f(x)g(x)]'=f'(x)g(x)+f(x)*g'(x)两边积分∫[f(x)g(x)]'dx=∫f'(x)g(x)dx+∫f(x)*g'(x)dxf(x)g(x)=∫g(x)df(x)+∫f(x)dg(x)所以∫f(x)dg(x)=f(x)g(x)-∫g(x)df(x)
