Question

Choose the correct answer in each of the question.
$\int \frac{x}{(x-1)(x-2)}dx$ equals
$\left(a\right)log\left|\frac{(x-1{)}^2}{x-2}\right|+C\left(b\right)log\left|\frac{(x-2{)}^2}{x-1}\right|+C\left(c\right)log\left|\right(\frac{x-1}{x-2}{)}^2|\left(d\right)log|\left(\right(x-1\left)\right(x-2\left)\right)|+C$

Answer 1

Let $\frac{x}{(x-1)(x-2)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}$
$\Rightarrow x=A(x-2)+B(x-1)........\left(i\right)$
On substituting x=1 and 2 in Eq. (i), we get A =-1 and B =2
$\therefore \frac{x}{(x-1)(x-2)}=-\frac{1}{(x-1)}+\frac{2}{(x-2)}\therefore \int \frac{x}{(x-1)(x-2)}dx=\int \frac{(-1)}{x-1}dx+\int \frac{2}{x-2}dx=-log|x-1|+2log|x-2|+C=-log|x-1|+log|x-2{|}^2+C=log\left|\frac{(x-2{)}^2}{x-1}\right|+C[\because logb-loga=log\frac{b}{a}]$
So, the option (b) is correct.