Evaluate:
#int_0^oo dx/((x+(x^2+1)^(1/2))^3)#
Substitute #x= tant#, #dx = sec^2tdt#.
Note that #x in (0,+oo) => t in (0,pi/2)#, so:
#int_0^oo dx/((x+(x^2+1)^(1/2))^3) = int_0^(pi/2) (sec^2tdt)/((tant+(tan^2t+1)^(1/2))^3)#
Use now the trigonometric identity:
#1+tan^2t = sec^2t#
and note that #sec t# is positive for #t in (0,pi/2)#, so:
#(1+tan^2t)^(1/2) = sect#
#int_0^oo dx/((x+(x^2+1)^(1/2))^3) = int_0^(pi/2) (sec^2tdt)/(tant+sect)^3#
Expand now:
#tant = sint/cost#
#sect = 1/cost#
#int_0^oo dx/((x+(x^2+1)^(1/2))^3) = int_0^(pi/2) 1/cos^2t (dt)/(sint/cost+1/cost)^3#
#int_0^oo dx/((x+(x^2+1)^(1/2))^3) = int_0^(pi/2) 1/cos^2t (dt)/((sint+1)^3/cos^3t)#
#int_0^oo dx/((x+(x^2+1)^(1/2))^3) = int_0^(pi/2) (costdt)/(sint+1)^3#
#int_0^oo dx/((x+(x^2+1)^(1/2))^3) = int_0^(pi/2) (d(sint+1))/(sint+1)^3#
#int_0^oo dx/((x+(x^2+1)^(1/2))^3) =-[1/(2(sint+1)^2)]_0^(pi/2) #
#int_0^oo dx/((x+(x^2+1)^(1/2))^3) = -1/8+1/2 = 3/8#
