Here ,
#I=int1/(x^(5/2)(1+x^2)^(1/4))dx#
Substitute #color(red)(x=1/t=>dx=-1/t^2dt#
So,
#I=int1/((1/t)^(5/2)(1+1/t^2)^(1/4))(-1/t^2)dt#
#=-intt^(5/2)/((t^2+1)^(1/4)/t^(1/2))*1/t^2dt#
#=-intt^(5/2+1/2-2)/(t^2+1)^(1/4)dt#
#:.I=-intt/(t^2+1)^(1/4)dt#
Substitute #color(blue)((t^2+1)^(1/4)=u=>t^2+1=u^4#
#=>2tdt=4u^3du=>tdt=2u^3du #
#I=-int(2u^3)/udu#
#=>I=-2intu^2du#
#=>I=-2[u^3/3]+c#
#=-2/3(u)^3+c#
Subst. back , #color(blue)(u=(t^2+1)^(1/4)#
#I=-2/3(t^2+1)^(3/4)+c#
Again subst. #color(red)(t=1/x#
#:.I=-2/3(1/x^2+1)^(3/4)+c#
#=>I=-2/3(1+x^2)^(3/4)/(x^2)^(3/4)+c#
#:.I=-(2(1+x^2)^(3/4))/(3x^(3/2))+c#