原式=∫√[9-(x+2)²]dx令x+2=3sinu, u在[-π/2, π/2]区间则dx=3cosudu原式=∫3cosu(3cosudu)=9∫(cos²u)du=4.5∫(1+cos2u)du=4.5(u+0.5sin2u)+C=4.5[arcsin(x+2)/3+(x+2)/3√(1-(x+2)²/9)]+C=9/2arcsin[(x+2)/3]+(x+2)/2√[9-(x+2)²]+C
