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inverse of 6/(x-1)+3/(x+1)

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partial fractions (2(1+x))x(x²+3x+2)

Solution

1x −1x+2

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Solution steps

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(2(1+x))x(x²+3x+2)

Expand 2(1+x)

2+2xx(x²+3x+2)

Factor x(x²+3x+2): x(x+1)(x+2)

=2x+2x(x+1)(x+2)

Create the partial fraction template using the denominator x(x+1)(x+2)

2+2xx(x+1)(x+2) =a₀x +a₁x+1 +a₂x+2

Multiply equation by the denominator

x(2x+2)(x+1)(x+2)x(x+1)(x+2) =a₀x(x+1)(x+2)x +a₁x(x+1)(x+2)x+1 +a₂x(x+1)(x+2)x+2

Simplify

2+2x=a₀(x+1)(x+2)+a₁x(x+2)+a₂x(x+1)

Solve the unknown parameters by plugging the real roots of the denominator: 0, −1, −2

For the denominator root 0: a₀=1

For the denominator root −1: a₁=0

For the denominator root −2: a₂=−1

a₀=1, a₁=0, a₂=−1

Plug the solutions to the partial fraction parameters to obtain the final result

1x +0x+1 +(−1)x+2

Simplify

1x −1x+2

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

inverse of 6/(x-1)+3/(x+1)

Solution

Solution steps

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