Question

求定积分 ∫1/x√lnx（1-lnx）dx 积分上限e^3/4 下限√e

做了一步 后边就不会了= =

Answer 1

令u = lnx，du = 1/x dx
当x = √e，u = 1/2
当x = e^(3/4)，u = 3/4
∫(√e~e^(3/4)) 1/[x√(lnx * (1 - lnx))] dx
= ∫(1/2~3/4) 1/√[u * (1 - u)] du
= ∫(1/2~3/4) 1/√(u - u²) du
= ∫(1/2~3/4) 1/√[- (u² - u + 1/4) + 1/4] du
= ∫(1/2~3/4) 1/√[1/4 - (u - 1/2)²] du
令u - 1/2 = (1/2)sinz，2u - 1 = sinz，2du = coszdz
当u = 1/2，0 = sinz => z = 0
当u = 3/4，1/2 = sinz => z = π/6
= ∫(0~π/6) 1/√(1/4 - 1/4 * sin²z) * (1/2)cosz dz
= ∫(0~π/6) dz
= π/6

Answer 2

∫(1/x)√lnx(1-lnx)dx
=∫√[1/4-(lnx-1/2)^2]d(lnx-1/2)
lnx-1/2=u
=∫√(1/4-u^2)du
=u√(1/4-u^2)+∫u^2du/√(1/4-u^2)
=u√(1/4-u^2)-∫√(1/4-u^2)du+∫(1/4)du/√(1/4)-u^2)
2∫(1/4-u^2)du=u√(1/4-u^2)+(1/4)∫d(2u)/√[1-(2u)^2]
=u√(1/4-u^2)+(1/4)arcsin(2u)+C1
∫√(1/4-u^2)du=(1/2)u√(1/4-u^2)+(1/8)arcsin(2u)+C

∫(1/x)√[lnx(l-lnx)]dx=(1/2)(lnx-1/2)√(lnx-lnx^2)+(1/8)arcsin(2lnx-1)
∫[√e, e^(3/4)] (1/x)√[lnx(1-lnx)]=(1/2)(1/4)√3/4+(1/8)arcsin(1/2)
=√3/32+π/48