∫1/(-x² + 6x - 10)dx = \na ln|-x² + 6x - 10|+c\nb -tan⁻¹(x - 3)+c\nc -tan⁻¹(x + 3)+c\nd tan⁻¹(x - 3)+c

Explanation:

Step1: Complete the square for the denominator

First, rewrite the denominator $-x^{2}+6x - 10$ as $-(x^{2}-6x + 10)$. Completing the square for $x^{2}-6x+10$, we have $x^{2}-6x + 10=(x - 3)^{2}+1$. So the integral becomes $\int\frac{1}{-( (x - 3)^{2}+1)}dx=-\int\frac{1}{(x - 3)^{2}+1}dx$.

Step2: Use the integral formula

Recall the integral formula $\int\frac{1}{u^{2}+1}du=\tan^{- 1}(u)+C$. Let $u=x - 3$, then $du=dx$. So $-\int\frac{1}{(x - 3)^{2}+1}dx=-\tan^{-1}(x - 3)+C$.

Answer:

B. $-\tan^{-1}(x - 3)+C$