How do you evaluate the integral #int dx/(x^3+x)#?

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Answer 1 · 2017-01-08T02:58:02

#int (dx)/(x^3+x) = 1/2 ln (x^2/(x^2+1))+C#

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#