Let $I =\int_1^2 \frac{x+3}{x(x+2)} d x$
Let $\frac{x+3}{x(x+2)}=\frac{A}{x}+\frac{B}{x+2}$
$\therefore x +3= A ( x +2)+ Bx$
Put $x=0$, we get
$ 3= A (2)+ B (0) $
$ \therefore A =\frac{3}{2}$\
Put $x+2=0$, i.e. $x=-2$, we get
$ -2+3= A (0)+ B (-2) $
$ \therefore 1=-2 B $
$ \therefore B =-\frac{1}{2}$
$\therefore \frac{x+3}{x(x+2)}=\frac{\left(\frac{3}{2}\right)}{x}+\frac{\left(-\frac{1}{2}\right)}{x+2}$
$\therefore I =\int_1^2\left[\frac{\left(\frac{3}{2}\right)}{x}+\frac{\left(-\frac{1}{2}\right)}{x+2}\right] d x $
$ =\frac{3}{2} \int_1^2 \frac{1}{x} d x-\frac{1}{2} \int_1^2 \frac{1}{x+2} d x $
$ =\frac{3}{2}[\log |x|]_1^2-\frac{1}{2}[\log |x+2|]_1^2$
$ =\frac{3}{2}(\log 2-\log 1)-\frac{1}{2}(\log 4-\log 3)$
$ =\frac{3}{2} \log 2-\frac{1}{2} \log 4+\frac{1}{2} \log 3 \quad \ldots[\because \log 1=0] $
$ \frac{1}{2}(3 \log 2-\log 4+\log 3)$
$=\frac{1}{2}(\log 8-\log 4+\log 3) $
$=\frac{1}{2} \log \left(\frac{8 \times 3}{4}\right)=\frac{1}{2} \log 6 .$
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