定积分的计算:√x⼀(1+x)dx在0到1的值

2025-01-24 09:54:21
推荐回答(2个)
回答1:

令t=√x,x=t²,t∈[0,1]
dx=2t
∫[0->1] √x/(1+x)dx=2∫[0->1] t²/(1+t²)dt
=2∫[0->1] 1-[1/(1+t²)]dt
=2-2arctant | [0->1]
=2-π/2

回答2:

1-ln2