求定积分∫(x+2)dx⼀根号2x+1.上限4,下限0.

需过程。谢谢!
2025-02-24 11:16:41
推荐回答(3个)
回答1:

∫[(x+2)/根号2x+1]dx
= ∫{[1/2(2x+1)+2/3]/根号2x+1}dx
=∫1/2(根号2x+1)dx+∫[3/2 /根号2x+1 ]dx
=1/6 *(2x+1)^(3/2)+3/2*(根号2x+1)
上限4,下限0.
得到 =22/3

回答2:

t=√(2x+1) 原积分式变为
1/2∫(t^2+3)dt,上限为3,下限为1
结果为22/3

回答3:

=∫(1/2*(2x+1)^(1/2)+3/2*(2x+1)^(-1/2))dx 上限4,下限0.
=1/3*(2x+1)^(3/2)+3(2x+1)^(1/2)
=9+9-1/3-3
=44/3