∫x√x+1dx的不定积分怎么求?

题目

∫x√x+1dx的不定积分怎么求?
x乘以(x+1)的1/2次方的不定积分怎么求?

答案

换元法.令t=√(x+1)
则x=t^2-1
dx=2tdt;
∫x√x+1dx=∫2t^2(t^2-1)dt
=∫(2t^4-2t^2)dt
=(2/5)t^5-(2/3)t^3+C
由t=√(x+1)
=(2/5)(x+1)^(5/2)-(2/3)(x+1)^(3/2)+C