Question

xln(x+根号1+x^2)的积分,用分部积分发, 不定积分，最好说一下此类题的思路，

Answer 1

简单分析一下，答案如图所示

Answer 2

∫xln[x+√(1+x²)]dx
=(1/2)∫ln[x+√(1+x²)]dx²
=(1/2)(x²ln[x+√(1+x²)]-∫x²dln[x+√(1+x²)])
=(1/2)(x²ln[x+√(1+x²)]-∫x²/√(1+x²)dx)
∫x²/√(1+x²)dx用三角
令x=tanθ,则θ=arctanx
∫x²/√(1+x²)dx
=∫sin²θ/cos³θdθ
=-∫sinθ/cos³θdcosθ
=(1/2)∫sinθd(1/cos²θ)
=(1/2)(sinθ/cos²θ-∫1/cosθdθ)
∫1/cosθdθ
=∫1/cos²θdsinθ
=∫1/(1-sin²θ)dsinθ
=(1/2)∫[1/(1+sinθ)+1/(1-sinθ)]dsinθ
=(1/2)ln|(1+sinθ)/(1-sinθ)|
然后一步一步回代就可以了,记得θ要换成x
过程比较麻烦,不易看懂,还要自己在草稿纸上算一遍