Explanation

(c)
Let $I = \int_{}^{}\frac{3x + 2}{(x - 2)^{2}(x - 3)}\ dx$ …(i)
Again, let
$\frac{3x + 2}{(x - 2)^{2}(x - 3)} = \frac{A}{(x - 2)} + \frac{B}{(x - 2)^{2}} + \frac{C}{(x - 3)}$ …(ii)
$\Rightarrow \ \ 3x + 2 = A(x - 2)(x - 3) + B(x - 3) + C(x - 2)^{2}$…(iii)
On putting the values of $x = 2,\ 3$ respectively, we get
$3 \times 2 + 2 = B(2 - 3)\ \ \ \ \Rightarrow \ \ \ B = - 8\$
and $3 \times 3 + 2 = C(3 - 2)^{2}\ \ \ \Rightarrow \ \ C = 11$
On equation the coefficient of $x^{2}$in Eq. (iii), we get
$0 = A + C\ \ \ \ \Rightarrow \ \ \ \ A = - 11$
On putting the values of $A,\ B$ and $C$ in Eq. (ii), we get
$\frac{3x + 2}{(x - 2)^{2}(x - 3)} = - \frac{11}{(x - 2)} - \frac{8}{(x - 2)^{2}} + \frac{11}{(x - 3)}$
$\therefore\ I = \int_{}^{}\left\lbrack - \frac{11}{(x - 2)} - \frac{8}{(x - 2)^{2}} + \frac{11}{(x - 3)}\ \right\rbrack dx$
$= - 11\log{(x - 2) + \frac{8}{(x - 2)} + 11\log{(x - 3)} + c}$
$= 11\ \log\left( \frac{x - 3}{x - 2} \right) + \frac{8}{(x - 2)} + c$