Question

∫arctan(1+√x)dx

Answer 1

∫ arctan(1+√x)dx
换元t=arctan(1+√x)，(tant -1)^2=x
=∫ t d(tant-1)^2
=t(tant-1)^2 - ∫ (tant-1)^2 dt
=t(tant-1)^2 - ∫ (sint-cost)^2/cos^2t dt
=t(tant-1)^2 - ∫ (1-2sintcost)/cos^2t dt
=t(tant-1)^2 - ∫ 1/cos^2t dt + 2∫ sint/cost dt
=t(tant-1)^2 - tant - 2∫ 1/cost d(cost)
=t(tant-1)^2 - tant - 2ln|cost| + C
=x*arctan(1+√x)-(1+√x)-2ln|cos(arctan(1+√x))| + C
有不懂欢迎追问

追问

=t(tant-1)^2 - ∫ (tant-1)^2 dt =t(tant-1)^2 - ∫ (sint-cost)^2/cos^2t dt看不懂

追答

因为
(tant-1)^2
=(sint/cost - 1)^2
=((sint-cost)/cost)^2
=(sint-cost)^2/cos^2t
所以
∫ (tant-1)^2 dt = ∫ (sint-cost)^2/cos^2t dt
有不懂欢迎追问

追问

+ 2∫ sint/cost dt是如何到 - 2∫ 1/cost d(cost)的?

追答

凑微分法：
∫ sint/cost dt
=∫ 1/cost * sintdt
=-∫ 1/cost * -sintdt
=-∫ 1/cost d(cost)
有不懂欢迎追问

Answer 2

令arctan(1+√x)=t，则x=(tant-1)^2
所以原式=∫t*2(tant-1)/cos^2(t)dt=∫2tsint/cos^3(t)dt-∫2t/cos^2(t)dt=-∫2t/cos^3(t)d(cost)-∫2td(tant)=-∫td(1/cos^2(t))-2∫td(tant)=-t/cos^2(t)+∫dt/cos^2(t)-2ttant+2∫tantdt=-t(1+tan^2(t))+tant-2ttant-2ln|cost|+C=-arctan(1+√x)*(x+2√x+2)+1+√x-2(1+√x)arctan(1+√x)-2ln√(x+2√x+2)+C