. $\int\limits_1^3 {\cfrac{{dx}}{{{x^2}\left( {x + 1} \right)}} = \cfrac{2}{3} + \log \cfrac{2}{3}}$

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

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

….(i)
Putting $x = 0$ in (i),

we get $1 = B\left( {0 + 1} \right)$ $\Rightarrow$ $B = 1$

Putting $x = - 1$ in (i),

we get $1 = C{\left( { - 1} \right)^2}$ $\Rightarrow$ $C = 1$

Comparing coefficients of ${x^2}$
on both sided of (i),

we get
$0 = A + C$ $\Rightarrow$ $A = - C$ $\Rightarrow$ $A = - 1$
$\therefore$
$\cfrac{1}{{{x^2}\left( {x + 1} \right)}} = - \cfrac{1}{x} + \cfrac{1}{{{x^2}}} + \cfrac{1}{{x + 1}}$

$\therefore$ $I = \int\limits_1^3 {\cfrac{{dx}}{{{x^2}\left( {x + 1} \right)}}} = \int\limits_1^3 {\left( { - \cfrac{1}{x} + \cfrac{1}{{{x^2}}} + \cfrac{1}{{x + 1}}} \right)dx}$

$= \left[ { - \log \left| x \right| + \cfrac{{{x^{ - 1}}}}{{ - 1}} + \log \left| {x + 1} \right|} \right]_1^3 = \left[ { - \cfrac{1}{x} + \log \left| {\cfrac{{x + 1}}{x}} \right|} \right]_1^3$

$= \left( { - \cfrac{1}{3} + 1} \right) + \log \cfrac{4}{3} - \log 2 = \cfrac{2}{3} + \log \left( {\cfrac{4}{3} \times \cfrac{1}{2}} \right) = \cfrac{2}{3} + \log \cfrac{2}{3}$