就是他的原函数
第1个回答 2010-09-07
分部积分法:
∫ f''(x)g(x) dx =∫ g(x) df'(x)
=f'(x)g(x) - ∫ f'(x) dg(x)
=f'(x)g(x) - ∫ f'(x)g'(x)dx
=f'(x)g(x) - ∫g'(x)df(x)
=f'(x)g(x) - [g'(x)f(x)-∫f(x)dg'(x)]
=f'(x)g(x) - g'(x)f(x)+∫f(x)g''(x) dx
∫ f''(x)g(x) dx =∫ g(x) df'(x)
=f'(x)g(x) - ∫ f'(x) dg(x)
=f'(x)g(x) - ∫ f'(x)g'(x)dx
=f'(x)g(x) - ∫g'(x)df(x)
=f'(x)g(x) - [g'(x)f(x)-∫f(x)dg'(x)]
=f'(x)g(x) - g'(x)f(x)+∫f(x)g''(x) dx