Let $\frac{1}{{\left( {{\rm{x}} + 2} \right){{\left(
{{\rm{x}} + 3} \right)}^2}}}$ = $\frac{{\rm{A}}}{{{\rm{x}} + 2}}$ +
$\frac{{\rm{B}}}{{{\rm{x}} + 3}}$ + $\frac{{\rm{C}}}{{{{\left( {{\rm{x}} + 3}
\right)}^2}}}$.
1 = A(x + 3)2 + B(x + 2)(x + 3) + C(x + 2).
Put x = -2, 1 = A.1
So, A = 1.
Or, x = -3, 1 = C.(-1)
So, C = 1
Or, x = 0, 1 = 9A + B.2.3 + C.2
Or, 6B = -6
So, B = -1
So, $\frac{1}{{\left( {{\rm{x}} + 2} \right){{\left(
{{\rm{x}} + 3} \right)}^2}}}$ = $\frac{1}{{{\rm{x}} + 2}} - \frac{1}{{{\rm{x}}
+ 3}} - \frac{1}{{{{\left( {{\rm{x}} + 3} \right)}^2}}}$.
Or, $\mathop \smallint \nolimits^ \frac{1}{{\left( {{\rm{x}}
+ 2} \right){{\left( {{\rm{x}} + 3} \right)}^2}}}$.dx = $\mathop \smallint
\nolimits^ \left\{ {\frac{1}{{{\rm{x}} + 2}} - \frac{1}{{{\rm{x}} + 3}} -
\frac{1}{{{{\left( {{\rm{x}} + 3} \right)}^2}}}} \right\}$.dx
= log(x + 2) – log(x + 3) + $\frac{1}{{{\rm{x}} + 3}}$ + c = log $\frac{{{\rm{x}} + 2}}{{{\rm{x}} + 3}}$ + $\frac{1}{{{\rm{x}} + 3}}$ + c.