Hello, r_maths!
$\displaystyle \text{Their answer adds up to: }\;\frac{x^3+3x^2+4x+3}{(x+1)(x+2)}$ . . . . obviously not the original fraction.Express $\displaystyle \frac{x^3+3x^2+5x+4}{(x+1)(x+2)}$ in partial fractions.
$\displaystyle \frac{x^3+3x^2+5x+4}{(x+1)(x+2)} \;=\;x + \frac{3x+4}{(x+1)(x+2)} \;=\;x + \frac{1}{x+1} + \frac{2}{x+2}\quad{\color{red}Right!} $
Book answer: .$\displaystyle x + \frac{1}{x+1} + \frac{1}{x+2}\quad{\color{red}Wrong!}$
But your answer checks out!
$\displaystyle \frac{x}{1}\cdot{\color{blue}\frac{(x+1)(x+2)}{(x+ 1)(x+2)}} + \frac{1}{x+1}\cdot{\color{blue}\frac{x+2}{x+2}} + \frac{2}{x+2}\cdot{\color{blue}\frac{x+1}{x+1}} $
. . $\displaystyle = \;\frac{x(x+1)(x+2) + (x+2) + 2(x+1)}{(x+1)(x+2)} $
. . $\displaystyle = \;\frac{x^3+3x^2+2x + x + 2 + 2x + 2}{(x+1)(x+2)} $
. . $\displaystyle = \;\frac{x^3+3x^2+5x+4}{(x+1)(x+2)} $ . . . . See?
$\displaystyle \frac{{x^3 + 3x^2 + 5x + 4}}
{{(x + 1)(x + 2)}} = \frac{{x^3 + 3x^2 + 2x + 3x + 4}}
{{(x + 1)(x + 2)}} = \frac{{x(x + 1)(x + 2) + (3x + 4)}}
{{(x + 1)(x + 2)}}$.
Now
$\displaystyle x + \frac{{3x + 4}}
{{(x + 1)(x + 2)}} = x + \frac{{(x + 2) + 2(x + 1)}}
{{(x + 1)(x + 2)}} = x + \frac{1}
{{x + 1}} + \frac{2}
{{x + 2}}$.
Hence $\displaystyle \frac{{x^3 + 3x^2 + 5x + 4}}
{{(x + 1)(x + 2)}} = x + \frac{1}
{{x + 1}} + \frac{2}
{{x + 2}}$.