What is expand (x-1)^{-1}-(x+3)^{-1} ?

The solution to expand (x-1)^{-1}-(x+3)^{-1} is 4/(x^2+2x-3)

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expand (x-1)^{-1}-(x+3)^{-1}

expand

(x - 1)^{- 1} - (x + 3)^{- 1}

\frac{4}{x ^{2} + 2 x - 3}

Solution steps

(x - 1)^{- 1} - (x + 3)^{- 1}

Apply exponent rule:

a^{- 1} = \frac{1}{a}

(x - 1)^{- 1} = \frac{1}{x - 1}

= \frac{1}{x - 1} - (x + 3)^{- 1}

Apply exponent rule:

a^{- 1} = \frac{1}{a}

(x + 3)^{- 1} = \frac{1}{x + 3}

= \frac{1}{x - 1} - \frac{1}{x + 3}

Least Common Multiplier of

x - 1, x + 3 : (x - 1) (x + 3)

Adjust Fractions based on the LCM

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

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

Expand

(x - 1) (x + 3) : x^{2} + 2 x - 3

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

Expand

x + 3 - (x - 1) : 4

= \frac{4}{x ^{2} + 2 x - 3}

(2 x^{2} - 5 π x - 3) (x^{2} - 2 x + e)

(x - 2) \cdot \frac{d}{d x} (1 - 4 4 - x^{2})

- \frac{2 l ( ( - 1 ) ^{n} - 1 )}{π ^{2} n ^{2}}

\frac{2 ( 1 + x ) y}{( 1 - 2 x - x ^{2} )}

\frac{3 ( 3 y - 1 )}{- 3 x + 2}

What is expand (x-1)^{-1}-(x+3)^{-1} ?
The solution to expand (x-1)^{-1}-(x+3)^{-1} is 4/(x^2+2x-3)