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

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

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

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

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