We can write the integral as:
#int (dx)/(x^3+x) = int (dx)/(x(1+x^2))#
now we can substitute:
#x=tant#
#dx= dt/cos^2t#
so we have:
#int (dx)/(x^3+x) = int (dt)/ (cos^2t tan t (1+tan^2t))#
and using the trigonometric identity:
#1+tan^2t = 1+sin^2t/cos^2t = (cos^2t+sin^2t)/cos^2t = 1/cos^2t#
#int (dx)/(x^3+x) = int (dt)/ (cos^2t tan t 1/cos^2t)=int (dt)/tant= int (costdt)/sint#
We can see that: #cost dt = d(sint)#, so:
#int (dx)/(x^3+x) = int (dsint)/sint = ln abs sin t +C#
To substitute back #x# we can note that:
#x=tant = sint/sqrt(1-sin^2t)#
#x^2= sin^2t/(1-sin^2t)#
#x^2(1-sin^2t)= sin^2t#
#x^2-x^2sin^2t= sin^2t#
#x^2=x^2sin^2t+ sin^2t#
#x^2=(x^2+1)sin^2t#
#x^2/(x^2+1) = sin^2t#
#sqrt(x^2/(x^2+1)) = abs sint#
so that finally:
#int (dx)/(x^3+x) = ln sqrt(x^2/(x^2+1)) +C = 1/2 ln (x^2/(x^2+1))+C#
