令u = lnx,du = 1/x dx
當(dāng)x = √e,u = 1/2
當(dāng)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
當(dāng)u = 1/2,0 = sinz => z = 0
當(dāng)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