Question

求不定积分 √(x^2+a^2)dx/x^4 (-1/3a^2)*[√（a^2+x^2)]^3/x^3+c

Answer 1

x=atanθ
dx=a(secθ)^2*dθ
x^2+a^2=a^2((tanθ)^2+1)=a^2*(secθ)^2
∫√(x^2+a^2)dx/x^4=
∫asecθ * a(secθ)^2 dθ/(a^4*(tanθ)^4)=
=1/a^2*∫cotθ*(cscθ)^3 dθ
u=cscθ
du=-cotθcscθ dθ
∫asecθ * a(secθ)^2 dθ/(a^4*(tanθ)^4)=
1/a^2*∫cotθ*(cscθ)^3 dθ=
-1/a^2∫u^2 du=
-1/(3a^2)*u^3+C=
-1/(3a^2)*(cscθ)^3+C=
x=atanθ=> tanθ=x/a => cscθ=(a^2+x^2)^(1/2)/(x)
∫√(x^2+a^2)dx/x^4=
-1/(3a^2)*(cscθ)^3+C=
-1/(3a^2)*(a^2+x^2)^(3/2)/x^3+C