Substitute #x=sqrt3 tant#, #dx = sqrt3sec^2t#:
#int dx/(x^2+3)^3 = sqrt3 int (sec^2t dt)/(3tan^2t+3)^3#
Use now the trigonometric identity:
#1+tan^2 t = sec^2t#
then:
#int dx/(x^2+3)^3 = int (sec^2t dt)/(3tan^2t+3)^3#
#int dx/(x^2+3)^3 = sqrt3/27int (sec^2t dt)/sec^6t#
#int dx/(x^2+3)^3 = sqrt3/27int ( dt)/sec^4t#
#int dx/(x^2+3)^3 = sqrt3/27int cos^4tdt#
Note now that using the identity: #2cos^2t = 1+cos2t#
#cos^4t = (cos^2t)^2 = (1+cos 2t)^2/4 = 1/4+1/2 cos2t + 1/4 cos^2 2t#
#cos^4t = 1/4+1/2 cos2t + 1/4 (1+cos 4t)/2#
#cos^4t = 3/8+1/2 cos2t + 1/8 cos 4t#
Then:
#int dx/(x^2+3)^3 = sqrt3/72 int dt + sqrt3/54 int cos2t dt+ sqrt3/216int cos 4t dt#
#int dx/(x^2+3)^3 = (sqrt3t)/72 + sqrt3/108 sin 2t dt+ sqrt3/864 sin 4t +C#
To undo the substitution use the parametric formulas:
#sin 2t = (2tant)/(1+tan^2t) = (2x/sqrt3)/(1+x^2/3) = (2sqrt3x)/(x^2+3)#
#sin 4t = 2sin2t cos2t = 2(2tant)/(1+tan^2t) (1-tan^2t)/(1+tan^2t)#
#sin 4t = -(4sqrt3x)/(x^2+3) (x^2-3)/(x^2+3)#
#sin 4t = -(4sqrt3x (x^2-3))/(x^2+3)^2#
Then:
#int dx/(x^2+3)^3 = sqrt3/72 arctan(x/sqrt3) + 1/18 x/(x^2+3)-1/72 (x(x^2-3))/(x^2+3)^2+C#
and summing the two rational functions:
#int dx/(x^2+3)^3 = sqrt3/72 arctan(x/sqrt3) + 1/72( (4x(x^2+3)- x(x^2-3))/(x^2+3)^2)+C#
#int dx/(x^2+3)^3 = sqrt3/72 arctan(x/sqrt3) + 1/72(( 3x^3+15x )/(x^2+3)^2)+C#
#int dx/(x^2+3)^3 = sqrt3/72 arctan(x/sqrt3) + 1/24(( x^3+5x )/(x^2+3)^2)+C#
