Question

The value of the integral $\int \frac{x}{(x-1{)}^2(x+2)}dx$
(where $m$ is an arbitrary constant)

• A. $\frac{2}{9}\ln \left|\frac{x-2}{x+4}\right|+\frac{1}{3(x-1)}+m$
• B. $\frac{2}{9}\ln \left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+m$
• C. $\frac{2}{3}\ln \left|\frac{x+1}{x+2}\right|-\frac{1}{3(x-1)}+m$
• D. $\frac{2}{3}\ln \left|\frac{x-1}{x+3}\right|-\frac{1}{3(x-1)}+m$

Answer 1

The correct option is B $\frac{2}{9}\ln \left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+m$

Let $\frac{x}{(x-1{)}^2(x+2)}=\frac{A}{(x-1)}+\frac{B}{(x-1{)}^2}+\frac{C}{(x+2)}$
$x=A(x-1)(x+2)+B(x+2)+C(x-1{)}^2$
By putting $x=1;x=-2$ and $x=0$, we obtain
$A=\frac{2}{9},B=\frac{1}{3},C=\frac{-2}{9}\therefore \frac{x}{(x-1{)}^2(x+2)}=\frac{2}{9(x-1)}+\frac{1}{3(x-1{)}^2}-\frac{2}{9(x+2)}\Rightarrow \int \frac{x}{(x-1{)}^2(x+2)}dx=\frac{2}{9}\int \frac{1}{(x-1)}dx+\frac{1}{3}\int \frac{1}{(x-1{)}^2}dx-\frac{2}{9}\int \frac{1}{(x+2)}dx=\frac{2}{9}\ln |x-1|+\frac{1}{3}\left(\frac{-1}{x-1}\right)-\frac{2}{9}\ln |x+2|+m=\frac{2}{9}\ln \left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+m$
Where $m$ is an arbitrary constant.