设f(x)是连续函数,且F(x)=∫xe−xf(t)dt,则F′(x)等于( )。−e−xf(e−x)−f(x)−e−xf(e−x)+f(x)e−xf(e−x)−f(x)e−xf(e−x)+f(x)A−e−xf(e−x)−f(x)正确答案B−e−xf(e−x)+f(x)Ce−xf(e−x)−f(x)De−xf(e−x)+f(x)