Evaluate the following integral:
$\int \frac{2 x+1}{(x-2)(x-3)} d x$
Let, $I=\int \frac{2 x+1}{(x-2)(x-3)} d x$
Now, let $\frac{2 x+1}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3}$
$\Rightarrow 2 x+1=A(x-3)+B(x-2)$
$\Rightarrow 2 x+1=(A+B) x-3 A-2 B$
Equating similar terms, we get,
$A+B=2$ and $3 A+2 B=-1$
So, $A=-5, B=7$
$\therefore \mathrm{I}=-5 \int \frac{\mathrm{dx}}{\mathrm{x}-2}+7 \int \frac{\mathrm{dx}}{\mathrm{x}-3}$
$\Rightarrow 1=-5 \log |x-2|+7 \log |x-3|+c$
$\Rightarrow 1=\log |x-2|^{-5}+\log |x-3|^{7}+c$
$\Rightarrow I=\log \left|\frac{(x-3)^{7}}{(x-2)^{5}}\right|+c$
Hence, $\int \frac{2 x+1}{(x-2)(x-3)} d x=\log \left|\frac{(x-3)^{7}}{(x-2)^{5}}\right|+c$
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