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

Factor the denominator: $x^2+3x+2 = (x+1)(x+2)$, then decompose:

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

• $x=-1:\ -2 = A(1) \Rightarrow A = -2$
• $x=-2:\ -4 = B(-1) \Rightarrow B = 4$

Integrate: …

The plan

The denominator factors into two distinct linear terms, so the fraction is a sum of two simple pieces, each integrating to a logarithm.

Step 1 — factor

Step 2 — decompose

Substitute the roots:

• $x=-1$: $\ -2 = A(1) \Rightarrow A = -2.$
• $x=-2$: $\ -4 = B(-1) \Rightarrow B = 4.$

Step 3 — integrate …

Method: Factor the quadratic denominator, then partial fractions

Use this for $\dfrac{\text{linear}}{ax^2+bx+c}$ when the quadratic factors over the reals — factor it into two distinct linear pieces and decompose into logarithms.

Steps

Step 1: Factor the denominator. Find the roots so that $ax^2+bx+c=a(x-r_1)(x-r_2)$ with $r_1\neq r_2$.

Step 2: Set up one constant per factor.

$\frac{px+q}{(x-r_1)(x-r_2)}=\frac{A}{x-r_1}+\frac{B}{x-r_2}.$ …

Common Mistakes

Mistake 1: Skipping the factoring step.

Why it's wrong: $\frac{2x}{x^2+3x+2}$ cannot be decomposed until you see $x^2+3x+2=(x+1)(x+2)$; leaving it as a quadratic blocks the cover-up method. Correct approach: always factor the denominator into linear pieces first.

Mistake 2: Sign slips at negative roots.

Why it's wrong: the roots are $x=-1,-2$; here $A=-2,\ B=4$, and mishandling the signs of $(x+1),(x+2)$ at those roots gives wrong constants. Correct approach: substitute $x=-1$ and $x=-2$ carefully into $2x=A(x+2)+B(x+1)$. …