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

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Answer 1 · 2016-06-17T06:16:52

#int2/(x(4x-1))dx=-2lnx+2ln(4x-1)=2ln((4x-1)/x)#

Partial fractions of #2/(x(4x-1))# are given by

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

#2/(x(4x-1))hArr(A(4x-1)+Bx)/(x(4x-1))# or

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

i.e. #A=-2# and #4A+B=0# i.e. #B=-4A=8#

Hence #2/(x(4x-1))=-2/x+8/(4x-1)# and

#int2/(x(4x-1))dx=int(-2/x+8/(4x-1))dx#

= #-2lnx+8xx1/4xxln(4x-1)#

= #-2lnx+2ln(4x-1)=2ln((4x-1)/x)#