分部积分法∫√(x^2+2x+5)dx＝∫√[(x+1)^2+4]d(x+1)＝√(x^2+2x+5)×(x+1)－∫(x+1)×(x+1)/√(x^2+2x+5)d(x+1)＝√(x^2+2x+5)×(x+1)－∫[√(x^2+2x+5)－4/√(x^2+2x+5)]dx＝√(x^2+2x+5)×(x+1)－∫√(x^2+2x+5)d...