1、∫ √(5-4x-x²) dx
=∫ √[9-(x+2)²] dx
令x+2=3sinu,則√[9-(x+2)²]=3cosu,dx=3cosudu
=∫ 9cos²u du
=(9/2)∫ (1+cos2u) du
=(9/2)u + (9/4)sin2u + C
=(9/2)u + (9/2)sinucosu + C
=(9/2)arcsin[(x+2)/3] + (9/2)[(x+2)/3]√[9-(x+2)²]/3 + C
=(9/2)arcsin[(x+2)/3] + (1/2)(x+2)√[9-(x+2)²] + C

2、∫ 1/√(4-x²)³ dx
令x=2sinu,則√(4-x²)³=8cos³u,dx=2cosudu
=∫ [1/(8cos³u)](2cosu) du
=(1/4)∫ sec²u du
=(1/4)tanu + C
=(1/4)x/√(4-x²) + C

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