MCQ
$\int \frac{x d x}{(x-1)(x-2)}$ equals
- A$\log \left|\frac{(x-1)^{2}}{x-2}\right|+C$
- B$\log |(x-1)(x-2)|+C$
- C$\log \left| {{{\left( {\frac{{x - 1}}{{x - 2}}} \right)}^2}} \right| + C$
- ✓$\log \left|\frac{(x-2)^{2}}{x-1}\right|+C$
$x=A(x-2)+B(x-1)$ .........$(1)$
Equating the coefficients of $x$ and constant, we obtain
$A+B=1$ and $-2 A-B=0$
$A=-1$ and $B=2$
$\therefore \frac{x}{(x-1)(x-2)}=-\frac{1}{(x-1)}+\frac{2}{(x-2)}$
$\Rightarrow \int \frac{x}{(x-1)(x-2)} d x=\int\left\{\frac{-1}{(x-1)}+\frac{2}{(x-2)}\right\} d x$
$=-\log |x-1|+2 \log |x-2|+C$
$=\log \left|\frac{(x-2)^{2}}{x-1}\right|+C$
Hence, the correct Answer is $D$.
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Let $F=4 x+6 y$ be the objective function. Maximum of $F-$ Minimum of $F=.....$
$ =\frac{1}{2024}, $ then $\alpha$ is equal to-
$[A]$ $\lim _{x \rightarrow 1^{-}} f(x)=0$
$[B]$ $\lim _{x \rightarrow 1^{-}} f(x)$ does not exist
$[C]$ $\lim _{x \rightarrow 1^{+}} f(x)=0$
$[D]$ $\lim _{x \rightarrow 1^{+}} f(x)$ does not exist