Integrals — Class 12 Maths Solution

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

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

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

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

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

we get
$2\left( 1 \right) - 3 = A\left( {1 + 1} \right)\left( {2 + 3} \right)$ $\Rightarrow$ $- 1 = A\left( 2 \right)\left( 5 \right)$ $\Rightarrow$ $A = - \cfrac{1}{{10}}$

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

we get
$- 2 - 3 = B\left( { - 1 - 1} \right)\left( { - 2 + 3} \right)$ $\Rightarrow$ $- 5 = B\left( { - 2} \right)\left( 1 \right)$ $\Rightarrow$ $B = \cfrac{5}{2}$
Putting $x = - \cfrac{3}{2}$ in $\left( i \right)$ ,

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

$\Rightarrow$ $- 6 = C\left( { - \cfrac{5}{2}} \right)\left( { - \cfrac{1}{2}} \right)$ $\Rightarrow$ $C = - 6 \times \cfrac{4}{5} = \cfrac{{ - 24}}{5}$

$\therefore$ $\cfrac{{2x - 3}}{{\left( {{x^2} - 1} \right)\left( {2x + 3} \right)}} = - \cfrac{1}{{10\left( {x - 1} \right)}} + \cfrac{5}{{2\left( {x + 1} \right)}} - \cfrac{{24}}{{5\left( {2x + 3} \right)}}$

$\Rightarrow$ $I = \int {\cfrac{{2x - 3}}{{\left( {{x^2} - 1} \right)\left( {2x + 3} \right)}}} dx$
$= - \cfrac{1}{{10}}\int {\cfrac{{dx}}{{x - 1}}} + \cfrac{5}{2}\int {\cfrac{{dx}}{{x + 1}} - \cfrac{{24}}{5}} \int {\cfrac{{dx}}{{2x + 3}}}$

$= - \cfrac{1}{{10}}\log \left| {x - 1} \right| + \cfrac{5}{2}\log \left| {x + 1} \right| - \cfrac{{24}}{{5 \times 2}}\log \left| {2x + 3} \right| + C$

$= \cfrac{5}{2}\log \left| {x + 1} \right| - \cfrac{1}{{10}}\log \left| {x - 1} \right| - \cfrac{{12}}{5}\log \left| {2x + 3} \right| + C$