问题补充:
求定积分∫ln[x+√(x²+1)] dx x属于[0,2]
答案:
答案是2ln(2 + √5) - √5 + 1,楼上算错
∫(0~2) ln[x + √(x² + 1)] dx
= { xln[x + √(x² + 1)] } |(0~2) - ∫(0~2) x dln[x + √(x² + 1)]
= 2ln(2 + √5) - ∫(0~2) x • 1/[x + √(x² + 1)] • [1 + x/√(x² + 1)] dx
= 2ln(2 + √5) - ∫(0~2) x/√(x² + 1) dx
= 2ln(2 + √5) - (1/2)∫(0~2) 1/√(x² + 1) d(x² + 1)
= 2ln(2 + √5) - (1/2) • 2√(x² + 1) |(0~2)
= 2ln(2 + √5) - √5 + 1
======以下答案可供参考======
供参考答案1:
∫ln[x+√(x²+1)] dx
=xln[x+√(x²+1)]|[0,2]- ∫[0,2]xdln[x+√(x²+1)]
=xln[x+√(x²+1)]|[0,2]- ∫[0,2]x/√(x²+1)dx
=2ln(2+√5)-2(x²+1)^(1/2)|(0,2)
=2ln(2+√5)-2√5+2