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

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

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Answer 1 · 2016-08-03T13:55:12

$\int \frac{4 x - 1}{x^{2} (x - 4)} d x = - \frac{15}{16} ln x - \frac{1}{4 x} + \frac{15}{16} ln (x - 4)$ $int(4x-1)/(x^2(x-4))dx=-15/16lnx-1/(4x)+15/16ln(x-4)$

Explanation:

Partial fractions of $\frac{4 x - 1}{x^{2} (x - 4)}$ $(4x-1)/(x^2(x-4))$ will be of type $\frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x - 4}$ $A/x+B/x^2+C/(x-4)$ i.e.

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

$\Leftrightarrow \frac{A x (x - 4) + B (x - 4) + C x^{2}}{x^{2} (x - 4)}$ $hArr(Ax(x-4)+B(x-4)+Cx^2)/(x^2(x-4))$ or

$\Leftrightarrow \frac{A x^{2} - 4 A x + B x - 4 B + C x^{2}}{x^{2} (x - 4)}$ $hArr(Ax^2-4Ax+Bx-4B+Cx^2)/(x^2(x-4))$ or

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

Hence $A + C = 0$ $A+C=0$ , $- 4 A + B = 4$ $-4A+B=4$ and $4 B = 1$ $4B=1$

i.e. $B = \frac{1}{4}$ $B=1/4$ , $- 4 A = 4 - \frac{1}{4} = \frac{15}{4}$ $-4A=4-1/4=15/4$ or $A = - \frac{15}{16}$ $A=-15/16$ and $C = \frac{15}{16}$ $C=15/16$ and

$\frac{4 x - 1}{x^{2} (x - 4)} \Leftrightarrow - \frac{15}{16 x} + \frac{1}{4 x^{2}} + \frac{15}{16 (x - 4)}$ $(4x-1)/(x^2(x-4))hArr-15/(16x)+1/(4x^2)+15/(16(x-4))$

Hence $\int \frac{4 x - 1}{x^{2} (x - 4)} d x \Leftrightarrow \int - \frac{15}{16 x} d x + \int \frac{1}{4 x^{2}} d x + \int \frac{15}{16 (x - 4)} d x$ $int(4x-1)/(x^2(x-4))dxhArrint-15/(16x)dx+int1/(4x^2)dx+int15/(16(x-4))dx$

= $- \frac{15}{16} ln x - \frac{1}{4 x} + \frac{15}{16} ln (x - 4)$ $-15/16lnx-1/(4x)+15/16ln(x-4)$

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

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