Decompose and integrate rational functions using partial fractions.

Integration by Partial Fractions — ∫dx/((x-1)(x+2))

Question

Evaluate:

\int \frac{dx}{(x-1)(x+2)}

Solution — Step by Step

We write the integrand as a sum of simpler fractions:

\frac{1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}

Why this form? Because each linear factor in the denominator gets its own constant numerator. This is the standard decomposition for non-repeated linear factors.

Multiply both sides by $(x-1)(x+2)$ :

1 = A(x+2) + B(x-1)

Now use the cover-up method — substitute the root of each factor:

Put $x = 1$ : $\quad 1 = A(3) + B(0) \Rightarrow A = \dfrac{1}{3}$
Put $x = -2$ : $\quad 1 = A(0) + B(-3) \Rightarrow B = -\dfrac{1}{3}$

\int \frac{dx}{(x-1)(x+2)} = \int \left(\frac{1/3}{x-1} - \frac{1/3}{x+2}\right) dx

= \frac{1}{3} \int \frac{dx}{x-1} \;-\; \frac{1}{3} \int \frac{dx}{x+2}

Both integrals are of the form $\int \frac{du}{u} = \ln|u| + C$ :

= \frac{1}{3}\ln|x-1| - \frac{1}{3}\ln|x+2| + C

Using $\ln a - \ln b = \ln\dfrac{a}{b}$ :

\boxed{\frac{1}{3} \ln\left|\frac{x-1}{x+2}\right| + C}

Why This Works

Partial fractions work because every rational function $\frac{P(x)}{Q(x)}$ — where degree of $P$ < degree of $Q$ — can be decomposed into simpler fractions whose integrals we already know. The key insight: $\int \frac{dx}{x-a} = \ln|x-a| + C$ is a direct consequence of the chain rule in reverse.

The cover-up method (substituting roots) is not a trick — it’s what you get when you isolate each constant algebraically. It just saves two minutes per problem, which adds up fast in JEE Main’s 3-minute-per-question pace.

CBSE marking scheme alert: CBSE 2025 Sample Paper awards 1 mark for correct decomposition, 1 mark for finding A and B, and 1 mark for the final integrated answer. Even if you make an arithmetic error finding A or B, you can recover marks if your method is shown clearly.

Alternative Method

Instead of the cover-up method, we can compare coefficients.

After clearing denominators: $1 = A(x+2) + B(x-1) = (A+B)x + (2A - B)$

Comparing coefficients on both sides:

Coefficient of $x$ : $A + B = 0$
Constant term: $2A - B = 1$

Adding both equations: $3A = 1 \Rightarrow A = \frac{1}{3}$ , then $B = -\frac{1}{3}$ .

Same result. This method is more systematic when you have repeated factors or quadratic factors in the denominator — the cover-up method doesn’t work cleanly there.

Common Mistake

Forgetting the negative sign on B. Students correctly find $A = \frac{1}{3}$ but then write the decomposition as $\frac{1/3}{x-1} + \frac{1/3}{x+2}$ (positive $B$ ). Check: $\frac{1/3}{x-1} + \frac{1/3}{x+2} = \frac{(x+2) + (x-1)}{3(x-1)(x+2)} = \frac{2x+1}{3(x-1)(x+2)}$ — clearly not $\frac{1}{(x-1)(x+2)}$ . Always verify by recombining your partial fractions before integrating.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next