令 √(1+x) = u, 1+x = u², dx = 2udu
∫x√(1+x)dx
=∫(u²-1)u×2udu
=2∫(u⁴- u²)du
=2[(u^5)/5 - u³/3] + C
=(2/5)[√(1+x)]^5 - ⅔[√(1+x)]³ + C
经验证,答案准确无误.
求不定积分（含根号）哪位高人帮忙解答一下?!∫x(根号(1+x))dx
令 √(1+x) = u, 1+x = u², dx = 2udu
∫x√(1+x)dx
=∫(u²-1)u×2udu
=2∫(u⁴- u²)du
=2[(u^5)/5 - u³/3] + C
=(2/5)[√(1+x)]^5 - ⅔[√(1+x)]³ + C
经验证,答案准确无误.