Question

Evaluate `int_1^2(x+3)/(x(x+2))  dx`

Answer 1

Let I =`int_1^2(x+3)/(x(x+2))  dx`

 Let`(x+3)/(x(x+2)) = A/x+B/(x+2)`

∴ x + 3 = A(x + 2) + Bx      ...(1)

Put x = 0 in (1), we get

3 = A(2) + B(0)

∴  A = `3/2`

Put x + 2 = 0, i.e. x = -2 in (1), we get

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

∴ 1 = -2B

∴ B = `-1/2`

∴ `(x+3)/(x(x+2))=((3/2))/x+((-1/2))/(x+2)`

∴ I = `int_1^2[((3/2))/x + ((-1/2))/(x+2)]dx`

`=3/2int_1^2 1/xdx-1/2int_1^2 1/(x+2)dx`

`=3/2[log|x|]_1^2-1/2[log|x+2|]_1^2`

`=3/2(log2-log1)-1/2(log4-log3)`

`=3/2log2-1/2log4+1/2log3        ...[because log1=0]`

`=1/2(3log2-log4+3)`

`=1/2(log8-log4+log3)`

`=1/2log((8xx3)/4)`

`=1/2log6`