Q.Find $\int \dfrac{3x - 2}{(x+1)^2 (x+3)}\, dx$

The repeated factor $(x+1)^2$ needs two terms:

Clear denominators: $3x-2=A(x+1)(x+3)+B(x+3)+C(x+1)^2$.

• $x=-1$: $-5=2B\Rightarrow B=-\tfrac52$.
• $x=-3$: $-11=4C\Rightarrow C=-\tfrac{11}{4}$.
• coefficient of $x^2$: $0=A+C\Rightarrow A=\tfrac{11}{4}$.

Integrate: …

Setting up the decomposition

The denominator $(x+1)^2(x+3)$ has a repeated linear factor $(x+1)$ and a distinct factor $(x+3)$. A repeated factor needs one term for each power:

Finding the constants

Clear denominators:

• Put $x=-1$: $3(-1)-2=-5=B(2)$, so $B=-\dfrac52$.
• Put $x=-3$: $3(-3)-2=-11=C(-3+1)^2=4C$, so $C=-\dfrac{11}{4}$.
• Compare $x^2$ coefficients: the left side has none, the right side has $A+C$, so $A+C=0\Rightarrow A=-C=\dfrac{11}{4}$.

Integrating term by term …

Method: Partial Fractions with a Repeated Linear Factor

Use this when the denominator contains a repeated factor like $(x+a)^2$: each power of the repeated factor needs its own term in the decomposition.

Steps

Step 1: Write one term per power of the repeated factor.

For $\dfrac{3x-2}{(x+1)^2(x+3)}$:

A repeated factor $(x+1)^2$ contributes both $\frac{A}{x+1}$ and $\frac{B}{(x+1)^2}$.

Step 2: Solve for the constants. …

Common Mistakes

Mistake 1: Writing only $\dfrac{A}{(x+1)^2}$ for the repeated factor.

Why it's wrong: a repeated factor needs a term for each power, so both $\frac{A}{x+1}$ and $\frac{B}{(x+1)^2}$ must appear. Correct approach: include one fraction per ascending power.

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