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

Step-by-step solution

1. 1

   Pick the template that a repeated factor demands. The denominator is the single linear factor $(x-1)$ raised to the power $2$, so the decomposition needs a term for every power from $1$ up to $2$:

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

   Using only the $(x-1)^2$ term leaves one unknown for a degree-1 numerator, which cannot work in general.

2. 2

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

   $2x-1=A(x-1)+B$

3. 3

   Expand and collect by powers of $x$.

   $2x-1=Ax+(B-A)$

4. 4

   Equate coefficients. Matching the $x$ terms gives $A=2$; matching the constants gives $B-A=-1$, so

   $B=A-1=1$

5. 5

   Cross-check with a shortcut substitution. Putting $x=1$ into $2x-1=A(x-1)+B$ kills the $A$ term immediately and gives $B=2(1)-1=1$ ✓ - a useful independent confirmation of the coefficient method.

6. 6

   State the result and verify numerically.

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

   At $x=3$: left $= frac{5}{4}=1.25$, right $= frac{2}{2}+ frac{1}{4}=1.25$ ✓. At $x=0.5$: left $=0$, right $=-4+4=0$ ✓.