1. We are asked to evaluate the integral $$\int \frac{1}{-x^2 + 4x - 5} \, dx.$$ 2. First, rewrite the denominator to a more recognizable form by completing the square: $$-x^2 + 4x - 5 = -(x^2 - 4x + 5).$$ Complete the square inside the parentheses: $$x^2 - 4x + 5 = (x^2 - 4x + 4) + 1 = (x - 2)^2 + 1.$$ So the denominator becomes: $$-(x - 2)^2 - 1 = -\left((x - 2)^2 + 1\right).$$ 3. Substitute this back into the integral: $$\int \frac{1}{-\left((x - 2)^2 + 1\right)} \, dx = -\int \frac{1}{(x - 2)^2 + 1} \, dx.$$ 4. Recall the standard integral formula: $$\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C.$$ Here, $a = 1$, so: $$-\int \frac{1}{(x - 2)^2 + 1} \, dx = -\tan^{-1}(x - 2) + C.$$ 5. Therefore, the integral evaluates to: $$-\tan^{-1}(x - 2) + C.$$ 6. Comparing with the given options, the correct answer is option B.