Question

求解不定积分

这个题咋做，不定积分的

Answer 1

简单计算一下，答案如图所示

Answer 2

rt

Answer 3

∫√(a²-x²)/x⁴dx
设x=asinu，dx=acosu du
∫√(a^2-x^2)/x^4 dx
=∫ [(acosu)/(asinu)^4] .[acosu du]
=(1/a^2) ∫ (cosu)^2/(sinu)^4 du
=(1/a^2) ∫ (cotu)^2(cscu)^2 du
=-(1/a^2) ∫ (cotu)^2. dcotu
=-[1/(3a^2)] (cotu)^3 + C
=-[1/(3a^2)] [√(a^2-x²)/x]^3+C

Answer 4

let
x=asinu
dx=acosu du
∫√(a^2-x^2)/x^4 dx
=∫ [(acosu)/(asinu)^4 ] (acosu du)
=(1/a^2) ∫ (cosu)^2/(sinu)^4 du
=(1/a^2) ∫ (cotu)^2. (cscu)^2 du
=-(1/a^2) ∫ (cotu)^2 dcotu
=-[1/(3a^2)] (cotu)^3 +C
=-[1/(3a^2)] [√(a^2-x^2)/x]^3 +C
where

x=asinu