Answer
$\frac{x^{2}+x+4}{(x+1)^{2}(x-1)}$
Work Step by Step
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:
$=\frac{x^{2}+x+4}{(x+1)^{2}(x-1)}$
