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 $x=-2$

$1=C$

$C=1$

Put $x=-1$

$-1=B$

$B=-1$

Equating coefficients of constants

$-1=2 A+2 B+C$

$-1=2 A-2+1$

$A=0$

Thus,

$I=0 \times \int \frac{d x}{x+1}+(-1) \int \frac{d x}{(x+1)^{2}}+\int \frac{d x}{x+2}$

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

$=\left(\frac{1}{(x+1)}\right)+\log |x+2|+C$