x^2⼀1+x在0到1上的定积分怎么求

2025-03-13 15:21:21
推荐回答(2个)
回答1:

x^2
= x(1+x) -x
=x(1+x) -(1+x) +1

∫(0->1) x^2/(1+x) dx
=∫(0->1) [ x-1 + 1/(1+x)] dx
= [(1/2)x^2 -x + ln|1+x|](0->1)
=(1/2) -1 +ln2
=ln2 - 1/2

回答2:

=2/3x^3/2+1/2x^2|0->1
=2/3+1/2-0
=7/6