Integrate using partial fractions: ∫dx/(x²-4)

Question

Evaluate: $\displaystyle\int \frac{dx}{x^2 - 4}$

Solution — Step by Step

x^2 - 4 = (x-2)(x+2)

This is a difference of squares. Now we can apply partial fractions.

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

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

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

Put $x = 2$ :

1 = A(4) + B(0) \Rightarrow A = \frac{1}{4}

Put $x = -2$ :

1 = A(0) + B(-4) \Rightarrow B = -\frac{1}{4}

So: $\dfrac{1}{x^2-4} = \dfrac{1/4}{x-2} - \dfrac{1/4}{x+2} = \dfrac{1}{4}\left(\dfrac{1}{x-2} - \dfrac{1}{x+2}\right)$

\int \frac{dx}{x^2-4} = \frac{1}{4}\int\frac{dx}{x-2} - \frac{1}{4}\int\frac{dx}{x+2}

= \frac{1}{4}\ln|x-2| - \frac{1}{4}\ln|x+2| + C

= \frac{1}{4}\ln\left|\frac{x-2}{x+2}\right| + C

Why This Works

Partial fractions convert a complex rational function into a sum of simpler ones, each of which integrates to a natural logarithm. The key insight is that $\int \frac{1}{x-a}dx = \ln|x-a| + C$ — this is the fundamental integral that makes partial fractions work.

The substitution method (putting $x = 2$ and $x = -2$ ) is faster than comparing coefficients because each substitution zeroes out all but one term.

Alternative Method — Using the Standard Result

There is a standard result for this type:

\int \frac{dx}{x^2 - a^2} = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| + C

Here $a = 2$ :

\int \frac{dx}{x^2-4} = \frac{1}{2 \times 2}\ln\left|\frac{x-2}{x+2}\right| + C = \frac{1}{4}\ln\left|\frac{x-2}{x+2}\right| + C

Same result. For JEE, memorising this standard form saves time.

Similarly: $\int \dfrac{dx}{a^2 - x^2} = \dfrac{1}{2a}\ln\left|\dfrac{a+x}{a-x}\right| + C$ . Notice the difference in sign and the order of terms in the fraction — don’t confuse the two.

Common Mistake

A frequent error: writing $\ln|x^2-4|/2 + C$ by treating the denominator as a single unit and using $\int f'/f = \ln|f|$ . This is wrong because $\frac{d}{dx}(x^2-4) = 2x \neq 1$ (the numerator is 1, not $2x$ ). The “direct log” method only works when the numerator is the derivative of the denominator. Here, we must use partial fractions.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the Standard Result

Common Mistake

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