Q.Integrate the following function: $\frac{x}{(x - 1)^2 (x + 2)}$

Repeated factor $(x-1)^2$ needs two terms:

Clearing, $x = A(x-1)(x+2) + B(x+2) + C(x-1)^2$:

• $x=1:\ 1 = 3B \Rightarrow B = \tfrac13$
• $x=-2:\ -2 = 9C \Rightarrow C = -\tfrac29$
• coeff. of $x^2:\ 0 = A+C \Rightarrow A = \tfrac29$

Integrate: …

The form

A squared linear factor contributes a term for each power up to its multiplicity:

Step 1 — clear denominators

Step 2 — solve for the constants

Substitute the roots to get $B$ and $C$ directly:

• $x=1$: $\ 1 = B(1+2) = 3B \Rightarrow B = \tfrac13.$
• $x=-2$: $\ -2 = C(-2-1)^2 = 9C \Rightarrow C = -\tfrac29.$

For $A$, compare the $x^2$ coefficients: the left side has none, the right side has $A + C$, so $A + C = 0 \Rightarrow A = \tfrac29.$

Step 3 — integrate term by term

• $\displaystyle \int \frac{2/9}{x-1}\,dx = \frac29\log|x-1|.$ …

Method: Partial Fractions with a Repeated Linear Factor

Use this whenever a proper rational function has a denominator carrying a repeated linear factor such as $(x-a)^2$ (or a higher power).

Steps

Step 1: Check the fraction is proper and factor the denominator.

The numerator degree must be less than the denominator degree; otherwise divide first. Identify every factor and its multiplicity.

Step 2: Write one term for EACH power of the repeated factor.

A factor $(x-a)^2$ contributes two terms — one for each power up to its multiplicity:

Step 3: Clear denominators and solve. …

Common Mistakes

Mistake 1: Writing only $\dfrac{B}{(x-1)^2}$ and dropping the $\dfrac{A}{x-1}$ term.

Why it's wrong: a repeated factor $(x-1)^2$ needs a term for every power from $1$ up to the multiplicity, so both $\dfrac{A}{x-1}$ and $\dfrac{B}{(x-1)^2}$ are required. Omitting $A$ makes the identity unsolvable. Correct approach: set up $\dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{C}{x+2}$.

Mistake 2: Integrating $\dfrac{1}{(x-1)^2}$ as a logarithm. …