$\cfrac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} - 4} \right)}}$

Let $I = \int {\cfrac{{5x\,dx}}{{\left( {x + 1} \right)\left( {{x^2} - 4} \right)}}dx}$

We write , $\cfrac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} - 4} \right)}} = \cfrac{{5x}}{{\left( {x + 1} \right)\left( {x + 2} \right)\left( {x - 2} \right)}}$

$\cfrac{{5x}}{{\left( {x + 1} \right)\left( {x + 2} \right)\left( {x - 2} \right)}} = \cfrac{A}{{x + 1}} + \cfrac{B}{{x + 2}} + \cfrac{C}{{x - 2}}$

$\Rightarrow$ $5x = A\left( {x + 2} \right)\left( {x - 2} \right) + B\left( {x + 1} \right)\left( {x - 2} \right) + C\left( {x + 1} \right)\left( {x + 2} \right)$

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

we get
$- 5 = A\left( { - 1 + 2} \right)\left( { - 1 - 2} \right)$ $\Rightarrow$ $A = \cfrac{5}{3}$

Putting $x = 2$ in (i),

we get
$10 = C\left( {2 + 1} \right)\left( {2 + 2} \right)$ $\Rightarrow$ $C = \cfrac{5}{6}$

$\therefore$ $\cfrac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} - 4} \right)}} = \cfrac{5}{{3\left( {x + 1} \right)}} - \cfrac{5}{{2\left( {x + 2} \right)}} + \cfrac{5}{{6\left( {x - 2} \right)}}$

$\therefore$ $I = \int {\cfrac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} - 4} \right)}}dx} = \cfrac{5}{3}\int {\cfrac{{dx}}{{x + 1}} - \cfrac{5}{2}\int {\cfrac{{dx}}{{x + 2}}} + \cfrac{5}{6}\int {\cfrac{{dx}}{{x - 2}}} }$

$= \cfrac{5}{3}\log \left| {x + 1} \right| - \cfrac{5}{2}\log \left| {x + 2} \right| + \cfrac{5}{6}\log \left| {x - 2} \right| + C$