Question

Find: 

`int (2x+3)/(x^2(x+3))  dx`

Answer 1

`I = int (2x+3)/(x^2(x+3))  dx`   ...(i)

`(2x+3)/(x^2 (x+3)) = A/x + B/x^2 + C/(x+3)`   ...(ii)

`(2x+3)/(x^2 (x+3)) = (A(x)(x+3)+B(x+3) + C (x^2))/(x^2 (x+3))`

2x + 3 = Ax(x + 3) + B(x + 3) + Cx²

Put x = 0

2(0) + 3 = A(0) (0 + 3) + B(0 + 3) + C(0)²

3 = 0 + 3B + 0

B = 1

Put x = −3

2(−3) + 3 = A(−3)(−3 + 3) + B(−3 + 3) + C(−3)²

−3 = 0 + 0 + 9C

`C = -1/3`

Put x = 1

2(1) + 3 = A(1) (1 + 3) + B(1 + 3) +C(1)²

5 = 4A + 4B + C

`5 = 4A + 4(1) +(-1/3)` 

`4A = 5-4+ 1/3`

`4A = 1+ 1/3`

`4A = 4/3`

`A = 1/3`

value of A, B and C put in eq (ii)

`(2x+3)/(x^2(x+3)) = 1/(3x)+1/x^2- 1/(3(x+3))`   ...(iii)

from eq (i) and eq (iii)

`I = int(1/(3x)+1/x^2-1/(3(x+3)))dx`

`I = 1/3 int1/xdx +int1/x^2 dx-1/3 int 1/(x+3) dx`

`I = 1/3 log x + x^-1/-1 - 1/3 log (x+3)+C`

`= 1/3 log x -1/x-1/3 log (x+3)+C`