Question

$\frac{x - 3}{x + 1} + \frac{ {x}^{2} + 6x - 1}{ {x}^{2} - 1 }$
find the sum
please give me answer fast 三....​

Answer 1

$\frac{x - 3}{x + 1} + \frac{ {x}^{2} + 6x - 1}{ {x}^{2} - 1}$

$\frac{x - 3}{x + 1} + \frac{ {x}^{2} + 6x - 1}{(x -1 )(x + 1)}$

$\frac{(x - 1)(x - 3) + {x}^{2} + 6x - 1}{(x - 1)(x + 1)}$

$\frac{ {x}^{2} - 3x - x + 3 + {x}^{2} + 6x - 1 }{ {x}^{2} - 1 }$

$\frac{2 {x}^{2} + 2x + 2}{ {x}^{2} - 1}$