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

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Answer 1 · 2016-08-06T08:21:22

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

Let #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((4A+B)x-A)/(x(4x-1))#

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

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

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

= #-2int(dx)/x+8int1/(4x-1)dx#

= #-2ln|x|+8(1/4ln(4x-1))+c#

= #-2ln|x|+2ln|4x-1|+c#