Which is equivalent to 2x−1−1x+1\frac{2}{x - 1} - \frac{1}{x + 1}x−12−x+11?Ax+3x2−1\frac{x + 3}{x^2 - 1}x2−1x+3check_circleBx−3(x−1)(x+1)\frac{x - 3}{(x - 1)(x + 1)}(x−1)(x+1)x−3C1x2−1\frac{1}{x^2 - 1}x2−11D3x+1x2−1\frac{3x + 1}{x^2 - 1}x2−13x+1ExplanationCommon denominator: 2(x+1)−(x−1)(x−1)(x+1)=2x+2−x+1x2−1=x+3x2−1\frac{2(x + 1) - (x - 1)}{(x - 1)(x + 1)} = \frac{2x + 2 - x + 1}{x^2 - 1} = \frac{x + 3}{x^2 - 1}(x−1)(x+1)2(x+1)−(x−1)=x2−12x+2−x+1=x2−1x+3.