怎么求这个定积分,不会求被积函数的原函数。

2025-06-26 09:07:39
推荐回答(1个)
回答1:

∫[0:1]y(1+y²)^(3/2)dy
=½∫[0:1](1+y²)^(3/2)d(1+y²)
=½·(2/5)(1+y²)^(5/2)|[0:1]
=(1/5)[(1+1²)^(5/2) -(1+0²)^(5/2)]
=(4√2 -1)/5