How do you find the antiderivative of #int 1/(x^2(1+x^2)) dx#?

Use partial fraction expansion:

#1/(x^2(x^2 + 1)) = A/x + B/x^2 + (Cx +D)/(x^2 + 1)#

#1 = A(x(x^2 + 1)) + B(x^2 + 1) + (Cx +D)(x^2)#

Let #x = 0#:

B = 1

#1 - (x^2 + 1) = A(x(x^2 + 1)) + (Cx +D)(x^2)#

Let x = 1:

#-1 = A(1(1^2 + 1)) + (C + D)(1^2)#

-1 = 2A + C + D

Let x = -1

#1 - (-1^2 + 1) = A(-1(-1^2 + 1)) + (C-1 +D)(-1^2)#

-1 = -2A - C + D

D = -1

A = C = 0

Check:

#1/x^2 - 1/(x^2 + 1) = #

#1/x^2(x^2 + 1)/(x^2 + 1) - 1/(x^2 + 1)(x^2)/(x^2) = #

#(x^2 + 1)/(x^2(x^2 + 1)) - (x^2)/(x^2(x^2 + 1)) = #

#1/(x^2(x^2 + 1)) # This checks.

#int1/(x^2(x^2 + 1))dx = int1/x^2dx - int1/(x^2 + 1)dx#

This method avoids partial fractions and uses a trig substitution.

#I=int1/(x^2(1+x^2))dx#

Let #x=tantheta#. This implies that #dx=sec^2thetad theta#. Thus:

#I=intsec^2theta/(tan^2theta(1+tan^2theta))d theta#

Since #1+tan^2theta=sec^2theta#:

#I=int1/tan^2thetad theta=intcot^2thetad theta#

Note that #cot^2theta=csc^2theta-1#:

#I=intcsc^2thetad theta-intd theta#

These are common integrals:

#I=-cottheta-theta+C#

Note that #tantheta=x#, so #cottheta=1/x# and #theta=arctanx#.

#I=-1/x-arctanx+C#