1. **State the problem:** Simplify the expression $$\frac{a + 1}{a^{2}-4} + \frac{1 - a^{2}}{(a-1)^{2} + 3}$$. 2. **Recall important formulas and rules:** - Difference of squares: $$a^{2} - 4 = (a-2)(a+2)$$. - Factorization: $$1 - a^{2} = (1 - a)(1 + a)$$. - Expand and simplify denominators carefully. 3. **Rewrite the expression using factorization:** $$\frac{a + 1}{(a-2)(a+2)} + \frac{(1 - a)(1 + a)}{(a-1)^{2} + 3}$$ 4. **Simplify the denominator of the second fraction:** $$(a-1)^{2} + 3 = (a^{2} - 2a + 1) + 3 = a^{2} - 2a + 4$$ 5. **Rewrite the second fraction:** $$\frac{(1 - a)(1 + a)}{a^{2} - 2a + 4}$$ 6. **Notice that $(1 - a) = -(a - 1)$, so:** $$(1 - a)(1 + a) = -(a - 1)(a + 1)$$ 7. **The expression becomes:** $$\frac{a + 1}{(a-2)(a+2)} - \frac{(a - 1)(a + 1)}{a^{2} - 2a + 4}$$ 8. **Find a common denominator:** The denominators are $(a-2)(a+2)$ and $a^{2} - 2a + 4$. Since these are different quadratics, the common denominator is their product: $$ (a-2)(a+2)(a^{2} - 2a + 4) $$ 9. **Rewrite each fraction with the common denominator:** $$\frac{(a + 1)(a^{2} - 2a + 4)}{(a-2)(a+2)(a^{2} - 2a + 4)} - \frac{(a - 1)(a + 1)(a-2)(a+2)}{(a-2)(a+2)(a^{2} - 2a + 4)}$$ 10. **Expand the numerators:** - First numerator: $$ (a + 1)(a^{2} - 2a + 4) = a^{3} - 2a^{2} + 4a + a^{2} - 2a + 4 = a^{3} - a^{2} + 2a + 4 $$ - Second numerator: $$ (a - 1)(a + 1)(a-2)(a+2) = (a^{2} - 1)(a^{2} - 4) = a^{4} - 4a^{2} - a^{2} + 4 = a^{4} - 5a^{2} + 4 $$ 11. **Combine the numerators:** $$ a^{3} - a^{2} + 2a + 4 - (a^{4} - 5a^{2} + 4) = -a^{4} + a^{3} + 4a^{2} + 2a $$ 12. **Final simplified expression:** $$ \frac{-a^{4} + a^{3} + 4a^{2} + 2a}{(a-2)(a+2)(a^{2} - 2a + 4)} $$ This is the simplified form of the original expression. **Answer:** $$ \frac{-a^{4} + a^{3} + 4a^{2} + 2a}{(a-2)(a+2)(a^{2} - 2a + 4)} $$