Question

如图这个不定积分有什么问题

用第一类换元法和直接带进去算出来怎么不一样

Answer 1

(1/2)∫ x^(1/2). x^(3/2) dx
=(1/2)∫ x^2 dx
=(1/6) x^3 + C

追问

你能不能用第一类换元法算一下

追答

(1/2)∫ x^(1/2). x^(3/2) dx
=(1/3)∫ x^(3/2) d(x^(3/2)
=(1/6) [x^(3/2)]^2 +C
=(1/6)x^3 +C

追问

不是1/2乘1/2吗怎么变成1/3了

追答

dx^(3/2) =(3/2)x^(1/2) dx

(1/2)∫ x^(1/2). x^(3/2) dx

=(1/3)∫  x^(3/2). [ 3/2)x^(1/2) dx]

=(1/3)∫  x^(3/2) dx^(3/2)
=(1/6) [x^(3/2)]^2  +C
=(1/6)x^3  +C

追问

不定积分x∧3/2等于多少

追答

u=x^(3/2)
∫x^(3/2) d x^(3/2)

=∫ u du
= (1/2) u^2 + C
= (1/2) [x^(3/2)]^2 + C
= (1/2) x^3 + C