`(2x+1)/(x-1)^2`

Since the denominator of the fraction is a repeated linear factor `(ax+b)^n` , its partial fraction decomposition is in the form:

`A_1/(ax+b) +A_2/(ax+b)^2+....+A_n/(ax+b)^n`

So,

`(2x+1)/(x-1)^2=A_1/(x-1)+A_2/(x-1)^2`

To determine the values of of A1 and A2, simplify the equation by multiplying both sides by the LCD of the three fractions.

`(x-1)^2((2x+1)/(x-1)^2)=(A_1/(x-1)+A_2/(x-1)^2)*(x-1)^2`

`2x+1=A_1(x-1)+A_2`       (Let this be EQ1.)

Then, assign a value to x, such that A1 will be eliminated. And only A2 remains in the equation.

To do so, let x=1.

`2(1)+1=A_1(1-1)+A_2`

`2+1=A_1*0+A_2`

`3=A_2`

Now that value of A2 is known, plug-in this to EQ1.

`2x+1=A_1(x-1)+A_2`

`2x+1=A_1(x-1)+3`

`2x+1-3=A_1(x-1)+3-3`

`2x-2=A_1(x-1)`

Then, assign another value to x.

Let it be x=0.

`2(0)-2=A_1(0-1)`

`0-2=A_1(-1)`

`-2=-A_1`

`2=A_1`

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

(For the second problem, kindly post it as a separate question in Homework Help.)