∫x√x+1dx (x根号x+1 dx）求不定积分._数学解答

∫x√x+1dx (x根号x+1 dx）求不定积分.

人气：369 ℃　时间：2020-05-25 08:27:57

解答

令√(x+1)=u,则x=u²-1,dx=2udu
原式=∫ (u²-1)*u*2udu
=2∫ (u^4-u²)du
=(2/5)u^5-(2/3)u³+C
=(2/5)(x+1)^(5/2)-(2/3)(x+1)^(3/2)+C