How do you write the partial fraction decomposition of the rational expression #1/(x^2(2x-1))#?

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Answer 1 · 2016-05-07T15:47:14

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

#1/(x^2(2x-1))# can be written in terms of partial fraction decomposition as

#1/(x^2(2x-1))hArrA/x+B/x^2+C/(2x-1)# and hence

#1/(x^2(2x-1))hArr(Ax(2x-1)+B(2x-1)+Cx^2)/(x^2(2x-1))# or

#1/(x^2(2x-1))hArr(2Ax^2-Ax+2Bx-B+Cx^2)/(x^2(2x-1))# or

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

#2A+C=0#, #2B-A=0# and #-B=1#

or #B=-1#, #A=2B=-2# and #C=-2A=4#

Hence, #1/(x^2(2x-1))hArr-2/x-1/x^2+4/(2x-1)#