Decompose (x + 1)/(x - 1)^2 into partial fractions.

Step-by-step solution

1. 1

   Choose the right template for a repeated factor. A factor $(x-1)^2$ needs a term for each power up to the multiplicity, not just the highest one:

   $\frac{x+1}{(x-1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}$

   Using only $\tfrac{B}{(x-1)^2}$ would be too few unknowns to match a degree-1 numerator.

2. 2

   Clear denominators. Multiply both sides by $(x-1)^2$:

   $x+1=A(x-1)+B$

3. 3

   Expand and group by powers of $x$.

   $x+1=Ax-A+B=Ax+(B-A)$

4. 4

   Equate coefficients. The coefficient of $x$ gives $A=1$; the constant term gives $B-A=1$, hence

   $B=1+A=2$

5. 5

   Write the decomposition and check.

   $\frac{x+1}{(x-1)^2}=\frac{1}{x-1}+\frac{2}{(x-1)^2},\qquad x\neq 1$

   Recombining: $\tfrac{(x-1)+2}{(x-1)^2}=\tfrac{x+1}{(x-1)^2}$ $\checkmark$. Numerically at $x=3$: left $=\tfrac44=1$, right $=\tfrac12+\tfrac24=1$ $\checkmark$.