Question

∫x√x+1dx的不定积分怎么求?

Answer 1

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