Substitute:
#x=tant#
#dx = (dt)/cos^2t#
#intx^3/(1+x^2)dx = int tan^3t/(1+tan^2t) dt/cos^2t#
Use the trigonometric identity:
#1+tan^2 t = 1/cos^2t#
#intx^3/(1+x^2)dx = int tan^3t/( 1/cos^2t) dt/cos^2t = int tan^3tdt#
Using the same identity again:
#int tan^3tdt = int tant tan^2t dt = int tant (1/cos^2t -1)dt =#
#= int tant d(tant) - int tant dt =1/2 tan^2 t - int sint /cost dt =#
#= 1/2tan^2t + int (d(cost))/cost = 1/2tan^2t + ln |cos t|#
To substitute back #x#, we have that:
#tant = x#
#cost= +-sqrt(1 /(1+tan^2t))= +-1/sqrt(1+x^2)#
So:
#intx^3/(1+x^2)dx = 1/2x^2-lnsqrt(1+x^2)#
