求不定积分√﹙x+1﹚－1/√﹙x+1）+1

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