这两道题的不定积分怎么做

2025-04-30 05:53:01

推荐回答（1个）

回答1：

(1)
let
x=siny
dx=cosy dy
∫ dx/(1-x^2)^(3/2)
=∫ dy/(cosy)^2
=∫ (secy)^2 dy
=tany + C
=x/√(1-x^2) + C

(3)

let

x=atany
dx=a(secy)^2 dy

∫dx/(x^2+a^2)^(3/2)
=(1/a^2)∫cosy dy
=(1/a^2) siny + C
=(1/a^2) x/√(x^2+a^2) + C

(5)
let
x= tany
dx=(secy)^2 dy

∫dx/[x^2.√(x^2+1) ]
=∫(secy)/(tany)^2 dy
=∫(cosy)/(siny)^2 dy
=∫dsiny/(siny)^2
=-1/siny + C
=-√(x^2+1)/x + C

(7)
∫dx/[(x+2)√(x+1) ]
=2∫d√(x+1) /(x+2)
=2√(x+1)/(x+2) -2∫√(x+1) dx
=2√(x+1)/(x+2) -(4/3)(x+1)^(3/2) +C