How do you integrate #int (4x-1)/( x^2(x-4))# using partial fractions?

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Answer 1 · 2016-08-03T13:55:12

#int(4x-1)/(x^2(x-4))dx=-15/16lnx-1/(4x)+15/16ln(x-4)#

Partial fractions of #(4x-1)/(x^2(x-4))# will be of type #A/x+B/x^2+C/(x-4)# i.e.

#(4x-1)/(x^2(x-4))hArrA/x+B/x^2+C/(x-4)# or

#hArr(Ax(x-4)+B(x-4)+Cx^2)/(x^2(x-4))# or

#hArr(Ax^2-4Ax+Bx-4B+Cx^2)/(x^2(x-4))# or

#(4x-1)/(x^2(x-4))hArr((A+C)x^2+x(-4A+B)-4B)/(x^2(x-4))#

Hence #A+C=0#, #-4A+B=4# and #4B=1#

i.e. #B=1/4#, #-4A=4-1/4=15/4# or #A=-15/16# and #C=15/16# and

#(4x-1)/(x^2(x-4))hArr-15/(16x)+1/(4x^2)+15/(16(x-4))#

Hence #int(4x-1)/(x^2(x-4))dxhArrint-15/(16x)dx+int1/(4x^2)dx+int15/(16(x-4))dx#

= #-15/16lnx-1/(4x)+15/16ln(x-4)#