∫10arctan(1+√x)dx的解是多少?∫arctan(1+√x)dx的解是多少？

Question

∫10arctan(1+√x)dx的解是多少?
∫arctan(1+√x)dx的解是多少？

Answer 1

令1+根号x=t
∫arctantd(t-1)^2=arctant*(t-1)^2-∫(t-1)^2darctant
=arctant*(t-1)^2-∫(t-1)^2/(1+t^2)dt
=arctant*(t-1)^2-∫(1-2t/(1+t^2))dt
=arctant*(t-1)^2-(t-ln|1+t^2|)+C
=arctant*(t-1)^2+ln|1+t^2|-t+C
代入t=根号x+1 即可 楼主应该会把