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

Question

1247 views around the world

You can reuse this answer
Creative Commons License

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

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

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

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

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