Question

1/(1-2x)^2 (1-3x) resolve into partial Fractions​

Answer 1

SOLUTION

TO DETERMINE

To resolve into partial Fractions

$\displaystyle \sf{ \frac{1}{ {( 1- 2x)}^{2}(1 - 3x) } }$

EVALUATION

$\displaystyle \sf{ \frac{1}{ {( 1- 2x)}^{2}(1 - 3x) } }$

$\displaystyle \sf{ = \frac{3(1 - 2x) - 2(1 - 3x)}{ {( 1- 2x)}^{2}(1 - 3x) } }$

$\displaystyle \sf{ = \frac{3(1 - 2x) }{ {( 1- 2x)}^{2}(1 - 3x) } - \frac{ 2(1 - 3x)}{ {( 1- 2x)}^{2}(1 - 3x) } }$

$\displaystyle \sf{ = \frac{3}{ ( 1- 2x)(1 - 3x) } - \frac{ 2}{ {( 1- 2x)}^{2} } }$

$\displaystyle \sf{ = 3 \times \frac{1}{ ( 1- 2x)(1 - 3x) } - 2 \times \frac{ 1}{ {( 1- 2x)}^{2} } }$

$\displaystyle \sf{ = 3 \bigg[ \frac{3(1 - 2x) - 2(1 - 3x)}{ ( 1- 2x)(1 - 3x) } \bigg] - \frac{2}{ {( 1- 2x)}^{2} } }$

$\displaystyle \sf{ = 3 \bigg[ \frac{3(1 - 2x) }{ ( 1- 2x)(1 - 3x) } - \frac{2(1 - 3x)}{ ( 1- 2x)(1 - 3x) } \bigg] - \frac{2}{ {( 1- 2x)}^{2} } }$

$\displaystyle \sf{ = 3 \bigg[ \frac{3 }{ (1 - 3x) } - \frac{2}{ ( 1- 2x) } \bigg] - \frac{2}{ {( 1- 2x)}^{2} } }$

$\displaystyle \sf{ = \frac{9}{ (1 - 3x) } - \frac{6}{ ( 1- 2x) } - \frac{2}{ {( 1- 2x)}^{2} } }$

$\displaystyle \sf{ = \frac{9}{ (1 - 3x) } + \frac{ - 6}{ ( 1- 2x) } + \frac{ - 2}{ {( 1- 2x)}^{2} } }$

FINAL ANSWER

$\displaystyle \sf{ \frac{1}{ {( 1- 2x)}^{2}(1 - 3x) } = \frac{9}{ (1 - 3x) } + \frac{ - 6}{ ( 1- 2x) } + \frac{ - 2}{ {( 1- 2x)}^{2} } }$

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