2道不定积分的题

1.∫(e^x-1)^1/2dx
设（e^x-1)^1/2=t
t^2=e^x-1, x=ln(t^2+1)
∫(e^x-1)^1/2dx =∫tdln(t^2+1)
=∫2t^2/(t^2+1)dt
=∫(2t^2+2-2)/(t^2+1)dt
=2-2∫1/(t^2+1)dt+C
=2-2arctan(t) +C
=2-2arctan[（e^x-1)^1/2] +C

2.
∫coslnxdx=(分部积分) xcoslnx-∫xdcoslnx
=xcoslnx+∫xsinlnx/x dx
=xcoslnx+∫sinlnx dx
=(分部积分) xcoslnx+xsinlnx-∫xdsinlnx
=xcoslnx+xsinlnx-∫coslnx dx +C
所以 2∫coslnx dx=xcoslnx+xsinlnx +C
∫coslnx dx=x(coslnx+sinlnx)/2 +C