Integrals — Class 12 Maths Solution

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

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

….(i)
Comparing coefficients of ${x^2}$ on both sides of (i),

we get
$0 = A + C$

Putting $x = - 2$ in $\left( i \right)$ ,

we get
$- 2 = C{\left( { - 2 - 1} \right)^2}$ $\Rightarrow$ $C = \cfrac{{ - 2}}{9}$

$\Rightarrow$ $A = - C = \cfrac{2}{9}$

Putting $x = 1$ in $\left( i \right)$ ,

we get
$1 = B\left( {1 + 2} \right)$ $\Rightarrow$ $B = \cfrac{1}{3}$
$\therefore$ $\int {\cfrac{x}{{{{\left( {x - 1} \right)}^2}\left( {x + 2} \right)}}} dx$

$= \int {\cfrac{2}{{9\left( {x - 1} \right)}}} dx + \int {\cfrac{1}{{3{{\left( {x - 1} \right)}^2}}}dx - \int {\cfrac{2}{{9\left( {x + 2} \right)}}} dx}$

$= \cfrac{2}{9}\log \left| {x - 1} \right| + \cfrac{1}{3} \times \cfrac{{{{\left( {x - 1} \right)}^{ - 1}}}}{{ - 1}} - \cfrac{2}{9}\log \left| {x + 2} \right| + C$

$= \cfrac{2}{9}\log \left| {\cfrac{{x - 1}}{{x + 2}}} \right| - \cfrac{1}{{3\left( {x - 1} \right)}} + C$