Question

Integrate the rational functions.
$\int \frac{x}{(x-1{)}^2(x+2)}dx.$

Answer 1

Let $\frac{x}{(x-1{)}^2(x+2)}=\frac{A}{x-1}+\frac{B}{(x-1{)}^2}+\frac{C}{x+2}....\left(i\right)$
$\Rightarrow x+A(x-1)(x+2)+B(x+2)+C(x-1{)}^2........(ii)$
On substituting x=1, -2 in Eq. (ii), we get
$1=B(1+2)\text{and}-2=C(-2-1{)}^2\Rightarrow B=\frac{1}{3}\text{and}C=\frac{-2}{9}$
On equating the coefficient of $x^2$ on both sides in Eq. (ii), we get
$0=A+C\Rightarrow A=-C=\frac{2}{9}$
$\therefore \int \frac{x}{(x-1{)}^2(x+2)}dx=\int \{\frac{\frac{2}{9}}{x-1}+\frac{\frac{1}{3}}{(x-1{)}^2}+\frac{(-\frac{2}{9})}{x+2}\}dx$ (from Eq.(i))
$\Rightarrow \int \frac{x}{(x-1{)}^2(x+2)}dx=\frac{2}{9}\int \frac{1}{x-1}dx+\frac{1}{3}\int \frac{1}{(x-1{)}^2}dx-\frac{2}{9}\int \frac{1}{x+2}dx=\frac{2}{9}log|x-1|+\frac{1}{3}\frac{(x-1{)}^{-2+1}}{(-2+1)}-\frac{2}{9}log|x+2|+C=\frac{2}{9}log\left|\frac{x-1}{x+2}\right|-\frac{1}{3}\left(\frac{1}{x-1}\right)+C[\because logb-loga=log(\frac{b}{a}\left)\right]$