Evaluate the following integral:

Question:

Evaluate the following integral:

$\int \frac{x}{(x-1)^{2}(x+2)} d x$

Solution:

$I=\int \frac{x}{(x-1)^{2}(x+2)} d x$

$\frac{\mathrm{x}}{(\mathrm{x}-1)^{2}(\mathrm{x}+2)}=\frac{\mathrm{A}}{\mathrm{x}-1}+\frac{\mathrm{B}}{(\mathrm{x}-1)^{2}}+\frac{\mathrm{c}}{\mathrm{x}+2}$

$x=A(x-1)(x+2)+B(x+2)+C(x-1)^{2}$

Put $x=-2$

$-2=9 c$

$C=-\frac{2}{9}$

Put $x=1$

$1=3 B$

$B=\frac{1}{3}$

Equating coefficients of constants

$0=-2 A+2 B+C$

$0=-2 A+2 * \frac{1}{3}-\frac{2}{9}$

$A=\frac{2}{9}$

Thus,

$I=\frac{2}{9} \int \frac{d x}{x-1}+\frac{1}{3} \int \frac{d x}{(x-1)^{2}}-\frac{2}{9} \int \frac{d x}{x+2}$

$I=\frac{2}{9} \log |x-1|+\frac{1}{3}\left(\frac{-1}{(x-1)}\right)-\frac{2}{9} \log |x+2|+C$

$=\frac{2}{9} \log \left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+C$

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