a=√(2x+1)
x=(a²-1)/2
dx=ada
原式=∫ada/[a(a²-1)/2]
=∫2da/(a²-1)
=∫2da/(a+1)(a-1)
=∫[1/(a-1)-1/(a+1)]da
=ln|a-1|-ln|a+1|+C
=ln|√(2x+1)-1|-ln[√(2x+1)+1]+C