What is #int (x^2-5x+4 ) / (x^3-2x +1 )#?

1 Answer
George C.
Jul 28, 2016

#int (x^2-5x+4)/(x^3-2x+1) dx#

#= ((5-9sqrt(5))/10)ln(abs(x+1/2-sqrt(5)/2)) + ((5+9sqrt(5))/10)ln(abs(x+1/2+sqrt(5)/2)) + C#

Explanation:

#(x^2-5x+4)/(x^3-2x+1)#

#=(color(red)(cancel(color(black)((x-1))))(x-4))/(color(red)(cancel(color(black)((x-1))))(x^2+x-1)#

#=(x-4)/(x^2+x-1)#

#=(x-4)/((x+1/2)^2-5/4)#

#=(x-4)/((x+1/2-sqrt(5)/2)(x+1/2+sqrt(5)/2))#

#=A/(x+1/2-sqrt(5)/2) + B/(x+1/2+sqrt(5)/2)#

#=(A(x+1/2+sqrt(5)/2)+B(x+1/2-sqrt(5)/2))/(x^2+x-1)#

#=((A+B)x + ((1/2+sqrt(5)/2)A+(1/2-sqrt(5)/2)B))/(x^2+x-1)#

Equating coefficients:

#{ (A+B=1), ((1/2+sqrt(5)/2)A+(1/2-sqrt(5)/2)B = -4) :}#

Subtract #1/2# of the first equation from the second to get:

#sqrt(5)/2(A-B) = -9/2#

Multiply both sides by #2/sqrt(5)# to find:

#A-B = -9/sqrt(5) = -(9sqrt(5))/5#

Add this to the first equation to find:

#2A = 1-(9sqrt(5))/5 = (5-9sqrt(5))/5#

Hence:

#A = (5-9sqrt(5))/10#

#B = (5+9sqrt(5))/10#

So:

#int (x^2-5x+4)/(x^3-2x+1) dx#

#= int ((5-9sqrt(5))/(10(x+1/2-sqrt(5)/2)) + (5+9sqrt(5))/(10(x+1/2+sqrt(5)/2))) dx#

#= ((5-9sqrt(5))/10)ln(abs(x+1/2-sqrt(5)/2)) + ((5+9sqrt(5))/10)ln(abs(x+1/2+sqrt(5)/2)) + C#

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