How do you find #int (x^2+1)/(x(x^2-1)) dx# using partial fractions?

Your goal is to "break" the fraction #(x^2 + 1)/(x*(x^2 - 1))# into several fractions.

First, do a complete factorization of the denominator: #x(x^2-1) = x(x+1)(x-1)#.

This means that you would like to find #A#, #B# and #C# so that the following holds:

#(x^2 + 1)/(x*(x+1)(x-1)) = A/x + B/(x+1) + C/(x-1)#

First, multiply the whole equation with #x*(x+1)(x-1)# in order to get rid of the denominator. You will get:

#x^2 + 1 = A(x+1)(x-1) + B(x-1)*x + C(x+1)*x#
#<=> x^2 + 1 = A (x^2 - 1) + B(x^2 - x) + C(x^2 + x)#

Now, the easiest way to go is to "gather" all #color(red)(x^2)# terms, all #color(blue)(x)# terms and all #color(green)("terms without "x)#.
Just so it's clear, your equation currently looks like this:

#color(red)(x^2) + color(blue)(0 * x) + color(green)(1) = A (color(red)(x^2) + color(blue)(0 *x) color(green)( - 1)) + B(color(red)(x^2) color(blue)(- x) + color(green)(0)) + C(color(red)(x^2) + color(blue)(x) + color(green)(0))#

This leads us to the following linear equation system:
the equation for #x^2#: #1 = A + B + C#
the equation for #x#: #0 = 0 - B + C#
the equation for #1#: #1 = -A#

#{ (1 = A + B + C), (0 = - B + C), (1 = -A):}#

Here, we can see immediately that #A = -1#. Pluging the value in the first equation leads us to the following system:

#{ (2 = B + C), (0 = - B + C), (-1 = A):}#

Now, you can add the first and the second equation which would lead you to #2 = 2C => C = 1#. Last but not least, it's easy to plug this value in a fitting equation to find out that #B= 1#.

This means that we have managed to compute the partial fractions:

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

This makes it easy to integrate:

#int (x^2 + 1)/(x*(x+1)(x-1)) "d"x = - int 1/x "d"x + int 1/(x+1) "d" x + int 1/(x-1) "d" x#

# = - ln | x | + ln | x + 1 | + ln | x - 1 | + c#.

Hope that this helped!