Integrals — Maths STD 12 Science — Question

$= \frac{{2x - 3}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {2x + 3} \right)}}$

$ = \frac{A}{{x - 1}} + \frac{B}{{x + 1}} + \frac{C}{{2x + 3}}$ ....(i)

$\Rightarrow $ 2x - 3 = A(x + 1)(2x + 3) + B(x - 1)(2x + 3) + C(x - 1)(x + 1)

$\Rightarrow $ 2x - 3 = A(2x² + 5x + 3) + B(2x² + x - 3) + C(x² - 1)

$\Rightarrow $ 2x - 3 = 2Ax² + 5Ax + 3A + 2Bx² + Bx - 3B + Cx² - C

Comparing coefficients of x²: 2A + 2B + C = 0 ...(ii)

Comparing coefficients of x : 5A + B = 2 .....(iii)

Comparing constants: 3A – 3B – C = –3 .....(iv)

On solving eq. (i), (ii) and (iii), we get $A = \frac{{ - 1}}{{10}},B = \frac{5}{2},C = \frac{{ - 24}}{5}$

Putting the values of A, B and C in eq. (i),

$\frac{{2x - 3}}{{\left( {{x^2} - 1} \right)\left( {2x + 3} \right)}}$

$= \frac{{\frac{{ - 1}}{{10}}}}{{x - 1}} + \frac{{\frac{5}{2}}}{{x + 1}} + \frac{{\frac{{ - 24}}{5}}}{{2x + 3}}$

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

$ = \frac{{ - 1}}{{10}}\log \left| {x - 1}+ \right|\frac{5}{2}\log \left| {x + 1} \right| - \frac{{24}}{5}\frac{{\log \left| {2x + 3} \right|}}{{2 }} + c$

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

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