根号下x^+2x+5的不定积分

网友回答

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)dx＋4×ln[x+1+√(x^2+2x+5)]
所以,∫√(x^2+2x+5)dx＝1/2×√(x^2+2x+5)×(x+1)＋2ln[x+1+√(x^2+2x+5)]＋C
======以下答案可供参考======
供参考答案1:
1/4*(2*x+2)*(x^2+2*x+5)^(1/2)+2*asinh(1/2*x+1/2)