Q.Integrate the following function: $\frac{3x - 1}{(x + 2)^2}$

Concept: Partial Fraction Decomposition — splitting a rational function into simpler fractions that integrate easily.

We have a repeated linear factor in the denominator, so we write:

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

Expand and compare coefficients:

So $A = 3$ and $2A + B = -1 \implies 6 + B = -1 \implies B = -7$.

Thus: …

Why Partial Fractions?

When you see a denominator like $(x+2)^2$, the natural instinct is to try a substitution $u = x+2$. That works, but partial fractions give a cleaner, more systematic approach — especially when the numerator is linear and the denominator is a repeated linear factor.

The key idea: any proper rational function (degree of numerator < degree of denominator) with a repeated linear factor $(ax+b)^n$ can be split into a sum of simpler fractions:

Here, $n=2$, so we expect two terms.

Step-by-step solution

1. Set up the decomposition

Since the denominator is $(x+2)^2$, we write:

where $A$ and $B$ are constants to be found.

2. Clear the denominator

Multiply both sides by $(x+2)^2$:

3. Solve for $A$ and $B$

Expand the right side:

Now compare coefficients of $x$ and the constant term:

• Coefficient of $x$: $3 = A$
• Constant term: $-1 = 2A + B$

Substitute $A = 3$ into the constant equation:

4. Rewrite the integral

Now the original integral becomes: …

Method: Partial Fractions for a Single Repeated Linear Factor

Use this when the entire denominator is a power of one linear factor, such as $(x+b)^2$, with a linear numerator on top.

Steps

Step 1: Write one term for each power up to the multiplicity.

Never use only the highest-power term — you'd lose $A$.

Step 2: Clear denominators.

Multiply through: $px+q = A(x+b) + B$.

Step 3: Solve for the constants. …

Common Mistakes

Mistake 1: Writing only $\dfrac{B}{(x+2)^2}$.

Why it's wrong: a repeated factor $(x+2)^2$ needs both $\dfrac{A}{x+2}$ and $\dfrac{B}{(x+2)^2}$; a single term cannot represent a linear numerator. Correct approach: use $\dfrac{A}{x+2}+\dfrac{B}{(x+2)^2}$, giving $A=3,\,B=-7$.

Mistake 2: Integrating $\dfrac{1}{(x+2)^2}$ as $\log|x+2|$. …