∫(1→4) x√(x² + 1) dx
= (1/2)∫(1→4) √(x² + 1) d(x² + 1)
= (1/2) * (2/3)(x² + 1)^(3/2) |(1→4)
= (17√17 - 2√2)/3
定积分∫(1→4) x(根号(x2+1)) dx
∫(1→4) x√(x² + 1) dx
= (1/2)∫(1→4) √(x² + 1) d(x² + 1)
= (1/2) * (2/3)(x² + 1)^(3/2) |(1→4)
= (17√17 - 2√2)/3