. $\cfrac{{{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}}$

: Let$I = \int {\cfrac{{{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}}} dx$

We write, $\cfrac{{{x^2} + x + 1}}{{{{\left( {x + 1} \right)}^2}\left( {x + 2} \right)}} = \cfrac{A}{{x + 1}} + \cfrac{B}{{{{\left( {x + 1} \right)}^2}}} + \cfrac{C}{{\left( {x + 2} \right)}}$

$\Rightarrow$ ${x^2} + x + 1 = A\left( {x + 1} \right)\left( {x + 2} \right) + B\left( {x + 2} \right) + C{\left( {x + 1} \right)^2}$

….(i)
Putting $x = - 1$ in (i),

we get
$1 - 1 + 1 = B\left( { - 1 + 2} \right)$ $\Rightarrow$ $B = 1$
Putting $x = - 2$ in (i),

we get
$4 - 2 + 1 = C{\left( { - 2 + 1} \right)^2}$ $\Rightarrow$ $C = 3$
Putting $x = 0$ in (i),

we get
$1 = 2A + 2B + C$ $\Rightarrow$ $1 = 2A + 2 + 3$ $\Rightarrow$ $A = - 2$

$\Rightarrow$ $I = \int {\left[ {\cfrac{{ - 2}}{{x + 1}} + \cfrac{1}{{{{\left( {x + 1} \right)}^2}}} + \cfrac{3}{{x + 2}}} \right]} dx$

$= - 2\log \left( {x + 1} \right) + \cfrac{{{{\left( {x + 1} \right)}^{ - 1}}}}{{ - 1}} + 3\log \left( {x + 2} \right) + C$
$= - 2\log \left( {x + 1} \right) - \cfrac{1}{{x + 1}} + 3\log \left( {x + 2} \right) + C$