求arctan根號下x的不定積分,
網友廻答:
- 設√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.
網友廻答:
- ∫arctan√xdx
=∫arctan√xd(x 1) 利用分部積分公式
=(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
網友廻答:
- ∫arctan√xdx
=xarctan√x-∫x*1/[1 (√x)^2]*1/2*1/√xdx
=xarctan√x-1/2*∫√x/(1 x)*dx (令√x=t,則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
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