Simplify the following expression: \[ \frac{1}{x+1}-\frac{2}{x^{2}+2 x+1}+\frac{3}{x^{2}-1} \] Question 2

To simplify the expression \(\frac{1}{x+1}-\frac{2}{x^{2}+2 x+1}+\frac{3}{x^{2}-1}\), start by recognizing that \(x^2 + 2x + 1 = (x+1)^2\) and \(x^2 - 1 = (x-1)(x+1)\). Substitute these into the expression: \[ \frac{1}{x+1} - \frac{2}{(x+1)^2} + \frac{3}{(x-1)(x+1)} \] Now, let’s find a common denominator, which is \((x+1)^2(x-1)\). Rewrite each term with this common denominator: 1. \(\frac{1}{x+1} = \frac{(x+1)(x-1)}{(x+1)^2(x-1)}\) 2. \(-\frac{2}{(x+1)^2} = \frac{-2(x-1)}{(x+1)^2(x-1)}\) 3. \(\frac{3}{(x-1)(x+1)} = \frac{3(x+1)}{(x-1)(x+1)(x+1)} = \frac{3(x+1)}{(x+1)^2(x-1)}\) Combining these three fractions under the common denominator: \[ \frac{(x+1)(x-1) - 2(x-1) + 3(x+1)}{(x+1)^2(x-1)} \] Now, expand the numerator: \[ (x^2 - 1) - 2x + 2 + 3x + 3 = x^2 + 2x + 1 = (x+1)^2 \] Putting it all together, we have: \[ \frac{(x+1)^2}{(x+1)^2(x-1)} = \frac{1}{x-1} \quad (\text{for } x \ne -1) \] Finally, the simplified expression is: \[ \frac{1}{x-1} \quad \text{(for } x \ne -1 \text{ and } x \ne 1\text{)} \]