∫x/(1+√x) dx
let
√x = (tany)^2
[1/(2√x)] dx = 2tany(secy)^2 dy
dx = 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∫ ( (secy)^6 - 3(secy)^4 +3(secy)^2 -1 ) tany dy
= 4[ (secy)^6/6 - 3(secy)^4/4 + 3(secy)^2/2 + ln|cosy| ] + C
= 4[ (1+√x)^3/6 - 3(1+√x)^2/4 + 3(1+√x)/2 - ln|1+√x| ] + C