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Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals5 Marks
Question
Integrate the rational function $\frac{3 x+5}{x^{3}-x^{2}-x+1}$
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Answer

We have, $\frac{3 x+5}{x^{3}-x^{2}-x+1}=\frac{3 x+5}{(x-1)^{2}(x+1)}$
Let $\frac{3 x+5}{(x-1)^{2}(x+1)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^{2}}+\frac{C}{(x+1)}$
$\Rightarrow 3x + 5 = A(x - 1)(x + 1) + B(x + 1) + C( x - 1)^2$
$\Rightarrow 3x + 5 = A(x^2 - 1) + B(x + 1) + C( x^2 + 1 - 2x) …(i)$
Substituting $x = 1,$ in equation $(i),$ we get,
$B = 4$
Equating the coefficients of $x^2$ and $x,$ we get,
$A + C = 0$
$B – 2C = 3$
$A=-\frac{1}{2}$ and $C=\frac{1}{2}$
Thus,
$\frac{3 x+5}{(x-1)^{2}(x+1)}=\frac{-1}{2(x-1)}+\frac{4}{(x-1)^{2}}+\frac{1}{2(x+1)}$
Now, $\int \frac{3 x+5}{(x-1)^{2}(x+1)} d x$ = $\int\left\{\frac{-1}{2(x-1)}+\frac{4}{(x-1)^{2}}+\frac{1}{2(x+1)}\right\} d x$
$= \frac{-1}{2} \int \frac{1}{(x-1)} d x+4 \int \frac{1}{(x-1)^{2}} d x+\frac{1}{2} \int \frac{1}{(x+1)} d x$
$=-\frac{1}{2} \log |x-1|+4\left(\frac{-1}{x-1}\right)+\frac{1}{2} \log |x+1|+C$
$= \frac{1}{2} \log \left|\frac{x+1}{x-1}\right|-\frac{4}{x-1}+C$

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