∫(0->1) 1/√(4 - x²) dx,用第二換元法
Let x = 2 * sinz,dx = 2 * cosz dz
x = 0,z = 0,x = 1,z = π/6
原式 = ∫(0->π/6) 1/(2cosz) * (2cosz) dz
= z |(0->π/6)
= π/6
∫(1->e) lnx dx,用分部積分法
= xlnx |(1->e) - ∫(1->e) x d(lnx)
= [(e)ln(e) - (1)ln(1)] - ∫(1->e) x * 1/x dx
= e - x |(1->e)
= e - (e - 1)
= 1