Mathematics

Evaluate :
$$\int {\dfrac{x}{{{{\left( {x\, - 1} \right)}^2}\left( {x + \,2} \right)}}} \,dx$$.


SOLUTION
$$\displaystyle \int \dfrac{x}{(x-1)^{2}(x+2)}dx $$
$$\displaystyle \dfrac{x}{(x-1)^{2}(x+2)} = \dfrac{A}{x-1}+\dfrac{B}{(x-1)^{2}}+\dfrac{C}{(x+2)} $$
$$ x = A(x-1)(x+2)+B(x+2)+C(x-1)^{2} $$
Equating coefficients  
$$ A+C  = 0 $$
$$ A+B-2C = 1 $$
$$ -2A+2B+C = 0 $$
On solving 
$$\displaystyle A = \dfrac{2}{9} C = \dfrac{-2}{9} B = \dfrac{1}{3} $$
$$\displaystyle \int \dfrac{x}{(x-1)^{2}(x+2)}dx = \frac{2}{9}\int \frac{1}{x-1}dx+\frac{1}{3}\int \frac{1}{(x-1)^{2}}dx-\frac{2}{9}\int \frac{1}{x+2}dx $$
$$\displaystyle I = \frac{2}{9}log|x-1|+\frac{1}{3}\frac{-1}{(x-1)}-\frac{2}{9}|x+2|+c $$
$$\displaystyle I = \frac{2}{9}log \frac{|x-1|}{|x+2|}-\frac{1}{3(x-1)}+C $$ 
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Subjective Medium Published on 17th 09, 2020
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