分母有理化
原式=∫[(x+1-2√(x+1)+1]/(x+1-1)] dx
=∫[1+1/x-2√(x+1)/x] dx
求∫√(x+1)/x dx
令a=√(x+1)
x=a²-1
dx=2ada
∫√(x+1)/x dx=∫a/(a²-1)*2ada
=2∫a²/(a²-1)da
=2∫(1+1/(a²-1)]da
=2∫[1+1/2[1/(a-1)-1/(a+1)]da
=2a+ln|(a-1)/(a+1)]+C
=2√(x+1)+ln|[√(x+1)-1]/[√(x+1)+1]+C
所以原式=x+ln|x|-4√(x+1)-2ln|[√(x+1)-1]/[√(x+1)+1]|+C