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

A repeated linear factor needs two terms:

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

• $x=-1$: $1=B$.
• $x=-2$: $3=C$.
• Coefficient of $x^2$: $1=A+C\Rightarrow A=-2$.

Integrate: …

Set-up

The fraction is proper (numerator degree $2<3$). The denominator has a repeated factor $(x+1)^2$ and a distinct factor $(x+2)$, so we need a term for each power of the repeated factor:

1. Solve for $A,B,C$

Multiply through by $(x+1)^2(x+2)$:

• Put $x=-1$: $1-1+1=B(1)\Rightarrow B=1$.
• Put $x=-2$: $4-2+1=C(1)\Rightarrow C=3$.
• Compare $x^2$ coefficients: $1=A+C\Rightarrow A=-2$.

(Quick check of the $x$-coefficient: $3A+B+2C=-6+1+6=1$ $\checkmark$.)

So

$\frac{x^2+x+1}{(x+1)^2(x+2)}=\frac{-2}{x+1}+\frac{1}{(x+1)^2}+\frac{3}{x+2}.$ …

Method: Partial fractions with a repeated linear factor

Use this for $\dfrac{P(x)}{(x-r)^{2}(x-s)}$: a factor raised to power $k$ needs one term for every power from $1$ up to $k$.

Steps

Step 1: Write terms for each power of the repeated factor.

Omitting the $\dfrac{B}{(x+1)^2}$ term is the standard mistake.

Step 2: Solve for the constants. …

Common Mistakes

Mistake 1: Omitting the $\dfrac{B}{(x+1)^2}$ term.

Why it's wrong: a squared factor $(x+1)^2$ needs terms for both the first and second power; leaving one out makes the decomposition unsolvable/incorrect. Correct approach: include $\dfrac{A}{x+1}+\dfrac{B}{(x+1)^2}$.

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

Why it's wrong: $\int(x+1)^{-2}\,dx=-\dfrac{1}{x+1}$, a power, not a log. Correct approach: use the power rule for the repeated term. …