1) half life:

t(1/2) = 1/k[A]o

2. Need to use method of partial fractions, then integrate:

1/(A-x)(B-X) = C/A-X + D/B-X

where C and D are constants to be evaluated. Continuing,

C(B-X) + D(A-X) =1

CB - CX +DA - DX = 1

Therefore

-CX - DX = 0, D = -C

and

CB + DA = 1

substituting -C for D above,

CB - CA =1

C = 1/B-A

hence, D= -1/B-A

Therefore

dx/([A]o - x)(**o - x) = **

dx C/([A]o - x) + dx D/(**o -x) =**

(1/(**o - [A]o))dx /([A]o-x) - (1/(****o - [A]o))dx/(****o - x) =**

(1/(**o - [A]o)) (dx /([A]o-x) - dx/(****o - x))**

now, the separated terms can be integrated via natural log.