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Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals5 Marks
Question
Integrate the rational function $\frac{{2x - 3}}{{\left( {{x^2} - 1} \right)\left( {2x + 3} \right)}}$
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Answer

$\frac{{2x - 3}}{{\left( {{x^2} - 1} \right)\left( {2x + 3} \right)}}$
$= \frac{{2x - 3}}{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {2x + 3} \right)}}$
$ = \frac{A}{{x - 1}} + \frac{B}{{x + 1}} + \frac{C}{{2x + 3}} ....(i)$
$\Rightarrow 2x - 3 = A(x + 1)(2x + 3) + B(x - 1)(2x + 3) + C(x - 1)(x + 1)$
$\Rightarrow 2x - 3 = A(2x^2 + 5x + 3) + B(2x^2 + x - 3) + C(x^2 - 1)$
$\Rightarrow 2x - 3 = 2Ax^2 + 5Ax + 3A + 2Bx^2 + Bx - 3B + Cx^2 - C$
Comparing coefficients of $x^2: 2A + 2B + C = 0 ...(ii)$
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$
$= \frac{{ - 1}}{{10}}\log \left| {x - 1}|+ \right|\frac{5}{2}\log \left| {x + 1} \right| - \frac{{12}}{5}\log \left| {2x + 3} \right| + c$
$= \frac{5}{2}\log \left| {x + 1} \right| - \frac{1}{{10}}\log \left| {x - 1} \right| - \frac{{12}}{5}\log \left| {2x + 3} \right| + c$

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