Suppose

$\frac{2x+1}{(x+2)(x-3)}=\frac{A}{x+2}+\frac{B}{x-3}….(1)$

Which implies $2x=1=A(x-3)+B(x+2)$

Now put, $x-3=0$, $x=3$

$2\times 3+1=A(0)+B(3+2)$

So, $B=\frac{7}{5}$

Now put $x+2=0$, $x=-2$

$-4+1=A(-2-3)+B(0)$

So,

$A=3/5$

From equation (1) we get,

$\frac{2x+1}{(x+2)(x-3)}=\frac{3}{5}\times \frac{1}{x+2}+\frac{7}{5}\times \frac{1}{x-3}$

\[\int \frac{2x-1}{(x+2)(x-3)}dx=\frac{3}{5}\int \frac{1}{x+2}dx+\frac{7}{5}\int \frac{1}{x-3}dx\]

\[\frac{3}{5}\log \left| x+2 \right|+\frac{7}{5}\log \left| x-3 \right|+c\]