Let I = \(\int\frac{(2x+1)}{(x+2)(x+3)}dx\),
Putting \(\frac{(2x+1)}{(x+2)(x+3)}\) = \(\frac{A}{x+2}+\frac{B}{x-3}\).......(1)
Which implies 2x=1 = A(x-3) + B(x+2)
Now put x-3=0, x=3
2 × 3 +1 = A(0 ) + B 3 + 2)
So B = 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,