Question

关于数学求解不定积分

有哪位大神可以帮忙做一下这个不定积分怎么做？过程详细一点，谢谢啦

Answer 1

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

Answer 2

Answer 3

请问a 是参数还是

追答