$\cfrac{{2x}}{{{x^2} + 3x + 2}}$

: Let $I = \int {\cfrac{{2x}}{{{x^2} + 3x + 2}}dx}$

We write, $\cfrac{{2x}}{{{x^2} + 3x + 2}} = \cfrac{{2x}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} = \cfrac{A}{{x + 1}} + \cfrac{B}{{x + 2}}$

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

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

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

Putting $x = - 2$ in (i),

we get
$2\left( { - 2} \right) = B\left( { - 2 + 1} \right)$ $\Rightarrow$ $B = 4$

$\therefore$ $\cfrac{{2x}}{{{x^2} + 3x + 2}} = \cfrac{{ - 2}}{{x + 1}} + \cfrac{4}{{x + 2}}$

$\Rightarrow$ $I = \int {\cfrac{{2x}}{{{x^2} + 3x + 2}}} dx = - 2\int {\cfrac{{dx}}{{x + 1}} + 4} \int {\cfrac{{dx}}{{x + 2}}}$

$= - 2\log \left| {x + 1} \right| + 4\log \left| {x + 2} \right| + C$