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

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Answer 1 · 2016-06-17T06:16:52

$\int \frac{2}{x (4 x - 1)} d x = - 2 ln x + 2 ln (4 x - 1) = 2 ln (\frac{4 x - 1}{x})$ $int2/(x(4x-1))dx=-2lnx+2ln(4x-1)=2ln((4x-1)/x)$

Partial fractions of $\frac{2}{x (4 x - 1)}$ $2/(x(4x-1))$ are given by

$\frac{2}{x (4 x - 1)} \Leftrightarrow \frac{A}{x} + \frac{B}{4 x - 1}$ $2/(x(4x-1))hArrA/x+B/(4x-1)$ or

$\frac{2}{x (4 x - 1)} \Leftrightarrow \frac{A (4 x - 1) + B x}{x (4 x - 1)}$ $2/(x(4x-1))hArr(A(4x-1)+Bx)/(x(4x-1))$ or

$\frac{2}{x (4 x - 1)} \Leftrightarrow \frac{x (4 A + B) - A}{x (4 x - 1)}$ $2/(x(4x-1))hArr(x(4A+B)-A)/(x(4x-1))$

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

Hence $\frac{2}{x (4 x - 1)} = - \frac{2}{x} + \frac{8}{4 x - 1}$ $2/(x(4x-1))=-2/x+8/(4x-1)$ and

$\int \frac{2}{x (4 x - 1)} d x = \int (- \frac{2}{x} + \frac{8}{4 x - 1}) d x$ $int2/(x(4x-1))dx=int(-2/x+8/(4x-1))dx$

= $- 2 ln x + 8 \times \frac{1}{4} \times ln (4 x - 1)$ $-2lnx+8xx1/4xxln(4x-1)$

= $- 2 ln x + 2 ln (4 x - 1) = 2 ln (\frac{4 x - 1}{x})$ $-2lnx+2ln(4x-1)=2ln((4x-1)/x)$

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