Integrals — Class 12 Maths Solution

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

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

$\Rightarrow$ $3x - 1 = A\left( {x + 2} \right) + B$

…(i)
Comparing coefficients of $x$ in (i), we get $A = 3$

Comparing the constant terms in (i),

we get
$2A + B = - 1$ $\Rightarrow$ $B = - 7$

$\therefore$ $\cfrac{{3x - 1}}{{{{\left( {x + 2} \right)}^2}}} = \cfrac{3}{{x + 2}} + \cfrac{{ - 7}}{{{{\left( {x + 2} \right)}^2}}}$

$\Rightarrow$ $I = \int {\cfrac{{3x - 1}}{{{{\left( {x + 2} \right)}^2}}}dx} = 3\int {\cfrac{{dx}}{{x + 2}}} - 7\int {\cfrac{{dx}}{{{{\left( {x + 2} \right)}^2}}}}$

$= 3\log \left| {x + 2} \right| - 7\cfrac{{{{\left( {x + 2} \right)}^{ - 1}}}}{{ - 1}} + C = 3\log \left| {x + 2} \right| + \cfrac{7}{{x + 2}} + C$