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0) Preparations needed?
As the power of your denominator (#x^3#) is bigger than the power of your numerator (#x^2#) and your denominator is already factorized, you can start with the partial fraction decomposition right away!
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1) Partial fraction decomposition
Your goal is to find #A#, #B# and #C# so that
#(x^2+1) / ((2x-1)(x+1)(x-1)) = A/(2x-1) + B/(x+1) + C/(x-1)#
... multiply both sides with #(2x-1)(x+1)(x-1)#....
#<=> x^2 + 1 = A(x+1)(x-1) + B(2x-1)(x-1) + C(2x-1)(x+1)#
... expand the terms at the right side...
#<=> x^2 + 1 = A * x^2 - A + B * 2x^2 - B* 3x + B + C * 2x^2 + C * x - C#
... "collect" the #color(red)(x^2)# terms, the #color(blue)(x)# terms and the #color(green)("constant")# terms...
#<=> color(red)(x^2) + color(blue)(0 * x) + color(green)(1) = color(red)(A x^2) + color(blue)(0 * Ax) color(green)(- A) + color(red)(2Bx^2) color(blue)(-3Bx) + color(green)( B) + color(red)(2Cx^2) + color(blue)(Cx) color(green)(-C)#
Now, to solve this equation, you can split it into three: one equation for #x^2#, one for #x# and one for the constant terms:
#{((I) color(white)(xxx) 1 = A + 2B + 2C color(white)(xxxxxxxxi) color(red)(x^2) " terms"),
((II) color(white)(xxi) 0 = -3B + C color(white)(xxxxxxxxxxx) color(blue)(x) " terms"),
((III) color(white)(xx) 1 = -A + B - C color(white)(xxxiii) color(green)("constant") " terms") :}#
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2) Solving linear equation system
To solve this system of linear equations, you can e.g. transform #(II)# as #C = 3B# and plug #3B# for #C# into #(I)# and #(III)# obtaining #(I')# and #(III')#:
#{ ( (I') color(white)(xx) 1 = color(white)(xx) A + 8B ),
( (III') color(white)(x) 1 = - A - 2B ) :}#
Adding those equation finally leads to #B = 1/3# and thus, the solution of the linear equation system is:
#A = - 5/3#, #color(white)(xx) B = 1/3#, #color(white)(xx) C = 1#.
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3) Result of the partial fraction decomposition
This means that your original fraction can be transformed as follows:
#(x^2+1) / ((2x-1)(x+1)(x-1)) = -5/3* 1/(2x-1) + 1/3* 1 /(x+1) + 1/(x-1)#
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4) Solving the integral
So, the last thing left to do is solving the integral!
#int (x^2+1) / ((2x-1)(x+1)(x-1)) "d" x#
#color(white)(xxxxxxxx) = int (-5/3* 1/(2x-1) + 1/3* 1 /(x+1) + 1/(x-1)) "d" x#
#color(white)(xxxxxxxx) = -5/3 int 1/(2x-1) "d"x + 1/3 int 1/(x+1) "d"x + int 1/(x-1) "d"x #
#color(white)(xxxxxxxx) = -5/3 ln |2x-1| * 1/2 + 1/3 ln |x+1| + ln |x-1| + c #
