$\int \frac{x^{2}+1}{(x-3)(x-2)} d x=P x+Q \log |x-3|+R \log |x-2|+c$, where $c$ is constant of integration,…
$\int \frac{x^{2}+1}{(x-3)(x-2)} d x=P x+Q \log |x-3|+R \log |x-2|+c$, where $c$ is constant of
integration, then the values of $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ are, respectively
$0,10,5$
$0,10,-5$
$1,10,5$
$1,10,-5$
Solution
$I=\int \frac{x^{2}+1}{(x-3)(x-2)} d x$
$=\int \frac{\left(x^{2}-5 x+6\right)+(5 x-5)}{(x-3)(x-2)} d x$
$\quad=\int \frac{x^{2}-5 x+6}{x^{2}-5 x+6} d x+5 \int \frac{x-1}{(x-3)(x-2)} d x$
Let $\frac{(x-1)}{(x-3)(x-2)}=\frac{A}{(x-3)}+\frac{B}{(x-2)}$
$\therefore(x-1)=A(x-2)+B(x-3)$
$\therefore A+B=1$ and $2 A+3 B=1 \Rightarrow A=2, B=-1$
$\therefore I=\int d x+5 \int\left[\frac{2}{(x-3)}-\frac{1}{(x-2)}\right] d x$
Thus $P=1, Q=10, R=-5$