Evaluate the following integral:

$I=\int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} d x$

$\frac{\mathrm{x}^{2}+\mathrm{x}+1}{(\mathrm{x}+1)^{2}(\mathrm{x}+2)}=\frac{\mathrm{A}}{\mathrm{x}+1}+\frac{\mathrm{B}}{(\mathrm{x}+1)^{2}}+\frac{\mathrm{c}}{\mathrm{x}+2}$

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

Put $x=-2$

$3=C$

$C=3$

Put $x=-1$

$1=B$

$B=1$

Equating coefficients of constants

$1=2 A+2 B+C$

$1=2 A+2+3$

$A=-2$

Thus,

$I=2 * \int \frac{\mathrm{dx}}{x+1}+(1) \int \frac{d x}{(x+1)^{2}}+3 \int \frac{d x}{x+2}$

$I=-2 \ln |x+1|-\left(\frac{1}{(x+1)}\right)+3 \ln |x+2|+C$