Now putting,

$\frac{{{x}^{2}}+x+1}{(x+2){{(x+1)}^{2}}}=\frac{A}{(x+2)}+\frac{B}{(x+1)}+\frac{C}{{{(x+1)}^{2}}}…….(1)$

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

Now put $x+1=0$

Therefore, $x=-1$

$A(0)+B(0)+C(-1+2)=1-1+1=1$

$C=1$

Now put $x+2=0$

Therefore, $x=-2$

$A{{\left( -2+1 \right)}^{2}}+B\left( 0 \right)+C\left( 0 \right)=4-2+1=3$

$A=3$

Equating the coefficient of ${{x}^{2}},A+B=1$

$3+B=1$

$B=-2$

From equation (1) we get,

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

So, $\int \frac{{{x}^{2}}+x+1}{\left( x+2 \right){{\left( x+1 \right)}^{2}}}dx=\int \frac{3}{(x+2)}dx-\int \frac{2}{(x+1)}dx+\int \frac{1}{{{(x+1)}^{2}}}dx$

$3\log \left| x+2 \right|-2\log \left| x+1 \right|-\frac{1}{1+x}+c$