Question

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

Answer 1

Let `(3x-2)/((x+1)^2(x+3)) = A/(x+1)+B/(x+1)^2 + C/(x+3)`

`3x - 2 = A(x+1)(x+3)+B(x+3)+C(x+1)^2`      ...(1)

Put x + 1 = 0, i.e. x = -1 in (1), we get

3(-1) - 2 = A(0)(2) + B(2) + C(0)

∴ -5 = 2B

∴ B = `-5/2`

Put x + 3 = 0, i.e. x = -3 in (1), we get

3(-3) - 2 = A(-2)(0) + B(0) + C(4)

∴ -11 = 4C

∴ C = `-11/4`

Put x = 0 in (1), we get

3(0) - 2 = A(1)(3) + B(3) + C(1)

∴ -2 = 3A + 3B + C

But `B = -5/2  "and"  C= -11/4`

∴ `-2 = 3A + 3(-5/2)-11/4`

∴ `3A = -2 + 15/2+11/4 = (-8+30+11)/4 = 33/4`

∴ `A = 11/4`

∴ `(3x-2)/((x+1)^2(x+3)) = (11/4)/(x+1)+ (-5/2)/(x+1)^2 + (-11/4)/(x+3)`

∴ I = `int[(11/4)/(x+1)+ (-5/2)/(x+1)^2 + (-11/4)/(x+3)]dx`

`=11/4int1/(x+1)dx-5/2int(x+1)^(-2)dx - 11/4int1/(x+3)dx`

`=11/4log|x+1|-5/2.((x+1)^-1)/-1 -11/4log|x+3|+c`

`=11/4log|(x+1)/(x+3)| + 5 /(2(x+1)) + c`