Integrate the rational function x (x-1)^ 2 (x+2) - Vidyadip

Question

Integrate the rational function $\frac{x}{(x-1)^{2}(x+2)}$

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Answer

Let $\frac{x}{(x-1)^{2}(x+2)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^{2}}+\frac{C}{(x+2)}$
$\Rightarrow$ x = A(x - 1)(x + 2) + B(x + 2) + C( x - 1)² …(i)
$\Rightarrow$ x = (A+C)x²+ (A+3B-2C)x - 2A+2B+C
Substituting x = 1 in equation (i), we get,
$B=\frac{1}{3}$
Equating the coefficients of x² and constant term, we get,
A + C = 0
-2A + 2B + C = 0
$A=\frac{2}{9}$ and $C=\frac{-2}{9}$
Thus,
$\frac{x}{(x-1)^{2}(x+2)}=\frac{2}{9(x-1)}+\frac{1}{3(x-1)^{2}}-\frac{2}{9(x+2)}$
$\Rightarrow$$\int \frac{x}{(x-1)^{2}(x+2)} d x=\int\left\{\frac{2}{9(x-1)}+\frac{1}{3(x-1)^{2}}-\frac{2}{9(x+2)}\right\} d x$
= $\frac{2}{9} \int \frac{1}{(x-1)} d x+\frac{1}{3} \int \frac{1}{(x-1)^{2}} d x-\frac{2}{9} \int \frac{1}{(x+2)} d x$
= $\frac{2}{9} \log |x-1|+\frac{1}{3}\left(\frac{-1}{x-1}\right)-\frac{2}{9} \log |x+2|+C$
= $\frac{2}{9} \log \left|\frac{x-1}{x+2}\right|+\frac{1}{3}\left(\frac{-1}{x-1}\right)+C$

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