Question

求解，有点急，帮帮忙！

Answer 1

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

Answer 2

换元法，令√(x+2)=t
则，x+2=t² ==> x=t²-2；dx=d(t²-2)=2tdt
原式=[2tdt/(t²-1)t]
=∫[2/(t²-1)]dt
=∫[1/(t-1)-1/(t+1)]dt
=ln|t-1|-ln|t+1|+C
=ln|(t-1)/(t+1)|+C
=ln|[√(x+2)-1]/[√(x+2)+1]|+C