一道数学积分题

∫√x√x√xdx

热心网友

∫√x√x√xdx=∫[x^(3/2)]dx=∫(2/5)d[x^(5/2)]=(2/5)x^(5/2) +C∫√[x√(x√x)]dx=∫x^(7/8)dx=∫(8/15)d[x^(15/8)]=(8/15)x^(15/8)+C

热心网友

问题疑是∫√[x√(x√x)]dx=∫{x*[x*x^(1/2)]^(1/2)}^(1/2)dx=∫x^(1/2)*x^(1/4)*x^(1/8)dx=∫x^(7/8)dx=8/15*x^(15/8)+C=8/15*x*x^(7/8)+C

热心网友

∫√x√x√xdx=∫[x^(3/2)]dx=∫(2/5)d[x^(5/2)]=(2/5)x^(5/2) +C∫√[x√(x√x)]dx=∫x^(7/8)dx=∫(8/15)d[x^(15/8)]=(8/15)x^(15/8)+C

热心网友

不知