∫x/(1+√x) dxlet√x = (tany)^2[1/(2√x)] dx = 2tany(secy)^2 dydx = 4(tany)^3(secy)^2 dy∫x/(1+√x) dx=∫[(tany)^4/(secy)^2] [4(tany)^3(secy)^2] dy=∫ 4(tany)^7 dy=4∫ [(secy)^2 -1)]^3 .tany dy=4∫ ( ...
求定积分∫x/(1+根号x)dx
∫x/(1+√x) dxlet√x = (tany)^2[1/(2√x)] dx = 2tany(secy)^2 dydx = 4(tany)^3(secy)^2 dy∫x/(1+√x) dx=∫[(tany)^4/(secy)^2] [4(tany)^3(secy)^2] dy=∫ 4(tany)^7 dy=4∫ [(secy)^2 -1)]^3 .tany dy=4∫ ( ...