求arctan根号下x的不定积分,

设√x=t,x=t^2,dx=2tdt,原式=2∫arctant*tdt
u=arctant,v'=t,u'=1/(1+t^2),v=t^2/2,
原式=(2*t^2/2)arctant-∫t^2dt/(1+t^2)
=t^2arctant-∫[（1+t^2)-1]dt/(1+t^2)
=t^2arctant-t-arctant+C
=xarctan√x-√x-arctan√x+C.

∫arctan√xdx
=∫arctan√xd(x+1) 利用分部积分公式
=(x+1)arctan√x - ∫(x+1)darctan√x
=(x+1)arctan√x - ∫(x+1)*[1/(1+(√x)^2)]d√x
=(x+1)arctan√x - ∫d√x
=(x+1)arctan√x - √x + C

∫arctan√xdx
=xarctan√x-∫x*1/[1+(√x)^2]*1/2*1/√xdx
=xarctan√x-1/2*∫√x/(1+x)*dx （令√x=t,则x=t^2,dx=2tdt）
=xarctan√x-1/2*∫t/(1+t^2)*2tdt
=xarctan√x-∫[1-1/(1+t^2)]dt
=xarctan√x-t+∫[1/(1+t^2)]dt
=xarctan√x-√x+arctan√x+C
=(x+1)arctan√x-√x+C