求e的根号下x-1次方的不定积分

2025-02-27 22:26:05
推荐回答(1个)
回答1:

设u=√(x-1),则x=u^2+1,dx=2udu,
∫e^√(x-1)*dx
=2∫ue^u*du
=2(u-1)e^u+c
=2[√(x-1)-1]e^√(x-1)+c.