∫上限1,下限0(x/(1+x的4次方)dx,求定积分
问题描述:
∫上限1,下限0(x/(1+x的4次方)dx,求定积分
答
∫上限1,下限0(x/(1+x的4次方)dx
=(1/2)∫上限1,下限0(1/(1+x的4次方)dx^2
=(1/2)arctanx^2|(0,1)
=π/8