∫arctan(1+√x)dx

∫ 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...=t(tant-1)^2 - ∫ (tant-1)^2 dt =t(tant-1)^2 - ∫ (sint-cost)^2/cos^2t dt看不懂因为(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+ 2∫ sint/cost dt是如何到 - 2∫ 1/cost d(cost)的?凑微分法：∫ sint/cost dt=∫ 1/cost * sintdt=-∫ 1/cost * -sintdt=-∫ 1/costd(cost)有不懂欢迎追问