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Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals3 Marks
Question
Integrate the rational function $\frac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} - 4} \right)}}$
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Answer

$\frac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} - 4} \right)}}$
$= \frac{{5x}}{{\left( {x + 1} \right)\left( {x + 2} \right)\left( {x - 2} \right)}}$
$= \frac{A}{{x + 1}} + \frac{B}{{x + 2}} + \frac{C}{{x - 2}} ...(i)$
$ \Rightarrow 5x = A(x + 2) + B(x + 1)(x - 2) + C(x + 1)(x + 2)$
$\Rightarrow 5x = A(x^2 4) + B(x^2 - x - 2) + C(x^2 + 3x + 2)$
$\Rightarrow 2x = Ax^2 - 4A + Bx^2 - Bx - 2B + Cx^2 + 3Cx + 2C$
Comparing coefficients of $x^2: A + B + C = 0 .......(ii)$
Comparing coefficients of $x: B + 3C= 5 .......(iii)$
Comparing constants: $4A +2B + 2C = 0 .......(iv)$
On solving eq. $(i), (ii)$ and $(iii),$ we get $A = \frac{5}{3},B = \frac{{ - 5}}{2},C = \frac{5}{6}$
Putting the values of $A, B$ and $C$ in eq. $(i),$
$\frac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} - 4} \right)}}$
$= \frac{{\frac{5}{3}}}{{x + 1}} + \frac{{\frac{{ - 5}}{2}}}{{\left( {x + 2} \right)}} + \frac{{\frac{5}{6}}}{{x - 2}}$
$\therefore \int \frac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} - 4} \right)}}dx$$= \frac{5}{3}\int {\frac{1}{{x + 1}}dx - \frac{5}{2}\int {\frac{1}{{\left( {x + 2} \right)}}dx + \frac{5}{6}} \int {\frac{1}{{x - 2}}} } dx$
$= \frac{5}{3}\log \left| {x + 1} \right| - \frac{5}{2}\log \left| {x + 2} \right| + \frac{5}{6}\log \left| {x - 2} \right| + c$

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