Decompose 1/(x²-1) into partial fractions

Question

Decompose $\dfrac{1}{x^2 - 1}$ into partial fractions.

Solution — Step by Step

x^2 - 1 = (x-1)(x+1)

This is a difference of squares. Both factors are distinct linear factors, so the partial fraction form has one term per factor.

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

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

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

This must hold for ALL values of $x$ .

Put $x = 1$ (makes the $B$ term vanish):

1 = A(1+1) + B(0) = 2A \Rightarrow A = \frac{1}{2}

Put $x = -1$ (makes the $A$ term vanish):

1 = A(0) + B(-1-1) = -2B \Rightarrow B = -\frac{1}{2}

\frac{1}{x^2-1} = \frac{1/2}{x-1} + \frac{-1/2}{x+1} = \frac{1}{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right)

Why This Works

Partial fractions decompose a rational function with a factorable denominator into a sum of simpler rational functions. The general rule: for distinct linear factors in the denominator, we get one $A/(x-a)$ term per factor. The constants are found by either substituting roots of each factor or by comparing coefficients.

The “substitution trick” (putting $x$ equal to each root) is the fastest method. It works because substituting the root of one factor zeroes out all terms except the one we want to find.

Alternative Method — Comparing Coefficients

From $1 = A(x+1) + B(x-1)$ , expand:

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

Comparing coefficients:

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

Adding: $2A = 1 \Rightarrow A = 1/2$ ; then $B = -1/2$ .

Same result. The substitution method is faster; the comparison method works even when substitution is messy.

Common Mistake

A very common error: writing $\frac{1}{x^2-1} = \frac{1}{x-1} \cdot \frac{1}{x+1}$ and then trying to integrate or simplify separately without partial fractions. This misses the point — you cannot split a sum of terms in the denominator by multiplying factors individually. You must use the partial fraction expansion.

The result $\frac{1}{x^2-1} = \frac{1}{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right)$ is a useful identity. Integrating: $\int \frac{dx}{x^2-1} = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C$ . This standard form appears frequently in JEE integration problems — worth memorising.

Question

Solution — Step by Step

Why This Works

Alternative Method — Comparing Coefficients

Common Mistake

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