Question

求定积分∫ln[x+√(x²+1）] dx x属于[0,2]

Answer 1

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

Answer 2

∫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

追问

dln[x+√(x²+1）] 怎么等于√(x²+1）dx啦？再详细点在
还有答案是2ln(2+√5)-√5+1哎

追答

∫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)-(x²+1)^(1/2)|(0,2)
=2ln(2+√5)-√5+1