How do you express $\frac{3}{(1 + x) (1 - 2 x)}$ $3/((1 + x)(1 - 2x))$ in partial fractions?

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Answer 1 · 2016-03-21T18:56:26

$\frac{3}{(1 + x) (1 - 2 x)} = \frac{1}{1 + x} + \frac{2}{1 - 2 x}$ $3/((1+x)(1-2x)) = 1/(1+x) +2/(1-2x)$

$\frac{3}{(1 + x) (1 - 2 x)} = \frac{A}{1 + x} + \frac{B}{1 - 2 x}$ $3/((1+x)(1-2x)) = A/(1+x) +B/(1-2x)$

$3 = A (1 - 2 x) + B ((1 + x)$ $3=A(1-2x)+B((1+x)$

$3 = A - 2 A x + B + B x$ $3=A-2Ax+B+Bx$

$A + B = 3, - 2 A + B = 0$ $A+B=3, -2A+B=0$

$B = 3 - A \to - 2 A + 3 - A = 0 \to - 3 A + 3 = 0$ $B=3-A->-2A+3-A=0->-3A+3=0$

$- 3 A = - 3, A = 1, B = 2$ $-3A=-3,A=1,B=2$

$\frac{3}{(1 + x) (1 - 2 x)} = \frac{1}{1 + x} + \frac{2}{1 - 2 x}$ $3/((1+x)(1-2x)) = 1/(1+x) +2/(1-2x)$

How do you express 3/((1 + x)(1 - 2x)) 3(1+x)(1−2x)3/((1 + x)(1 - 2x)) in partial fractions?