∫x√x+1dx (x根号x+1 dx)求不定积分.

问题描述:

∫x√x+1dx (x根号x+1 dx)求不定积分.

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