How do you integrate $\int \frac{4 x - 2}{3 {(x - 1)}^{2}}$ $int (4x-2) /( 3(x-1)^2)$ using partial fractions?

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How do you integrate $\int \frac{4 x - 2}{3 {(x - 1)}^{2}}$ $int (4x-2) /( 3(x-1)^2)$ using partial fractions?

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Answer 1 · 2016-05-12T16:32:05

$\int \frac{4 x - 2}{3 {(x - 1)}^{2}} d x = \frac{4}{3} ln (x - 1) - \frac{4}{3 (x - 1)} + c$ $int(4x-2)/(3(x-1)^2)dx=4/3ln(x-1)-4/(3(x-1))+c$

Explanation:

Let us first find partial fractions of $\frac{4 x - 2}{3 {(x - 1)}^{2}}$ $(4x-2)/(3(x-1)^2)$ and for this let

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

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

Hence $A = 4$ $A=4$ and $B - A = - 2$ $B-A=-2$ i.e. $B = 4 - 2 = 2$ $B=4-2=2$

Hence $\int \frac{4 x - 2}{3 {(x - 1)}^{2}} d x = \frac{1}{3} \int [\frac{2}{x - 1} + \frac{2}{{(x - 1)}^{2}}] d x$ $int(4x-2)/(3(x-1)^2)dx=1/3int[2/(x-1)+2/(x-1)^2]dx$

= $\frac{2}{3} \int \frac{2}{x - 1} d x + \frac{2}{3} \int \frac{2}{{(x - 1)}^{2}} d x + k$ $2/3int2/(x-1)dx+2/3int2/(x-1)^2dx+k$

= $\frac{4}{3} ln (x - 1) - \frac{4}{3 (x - 1)} + c$ $4/3ln(x-1)-4/(3(x-1))+c$

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How do you integrate $\int \frac{4 x - 2}{3 {(x - 1)}^{2}}$ $int (4x-2) /( 3(x-1)^2)$ using partial fractions?

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Explanation:

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