=∫1/(x+1)√(x²+x)dxx²+x=(x+1/2)²－(1/2)²三角换元脱根号令x=secu/2－1/2=∫(0.arccos(1/3))1/(secu/2+1/2)(tanu/2)d(secu/2－1/2)=2∫secu/(secu+1)du=2∫1/(cosu+1)du=∫sec²(u/2)du=2tan(u/2)
