## Precalculus: Mathematics for Calculus, 7th Edition

$\frac{x^{2}+x+4}{(x+1)^{2}(x-1)}$
Apply difference of squares rule to $(x+1)^{2}$: $(x+1)(x-1)$ Find the LCD for the three fractions. That is, $(x+1)^{2}(x-1)$. Adjust the fractions based on the LCD: $=\frac{(x+1)(x-1)}{(x+1)^{2}(x-1)}-\frac{2(x-1)}{(x+1)^{2}(x+1)}+\frac{3(x+1)}{(x+1)^{2}(x-1)}$ Combine the fractions: $=\frac{(x+1)(x-1)-2(x-1)+3(x+1)}{(x+1)^{2}(x-1)}$ Expand the numerator: $=\frac{x^{2}-1-2x+2+3x+3}{(x+1)^{2}(x-1)}$ Collect the like terms: $=\frac{x^{2}+x+4}{(x+1)^{2}(x-1)}$