Q.Integrate the following function: $\frac{1}{x^4 - 1}$

Concept: Partial Fraction Decomposition – factor the denominator into linear and quadratic factors, then integrate each term separately.

First, factor the denominator:

Set up the partial fractions:

Solving (by clearing denominators and comparing coefficients) gives:

Now integrate term by term: …

The key to integrating rational functions like $\frac{1}{x^4 - 1}$ is partial fraction decomposition. The idea is simple: a complicated fraction can be broken into a sum of simpler fractions, each of which is easy to integrate. Here, the denominator factors nicely into linear and irreducible quadratic factors, so we can split the fraction into pieces that integrate to logarithms and an inverse tangent.

Let’s work through it step by step.

1. Factor the denominator completely.

   $x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x-1)(x+1)(x^2+1)$.

   The factors are: two distinct linear factors $(x-1)$ and $(x+1)$, and one irreducible quadratic factor $(x^2+1)$ (it has no real roots).

2. Set up the partial fraction form.

   For each linear factor, we assign a constant numerator. For the irreducible quadratic, we assign a linear numerator (since the denominator is degree 2). So we write:

3. Clear denominators. Multiply both sides by $(x-1)(x+1)(x^2+1)$:

4. Solve for $A$, $B$, $C$, $D$.

   We can use a mix of substitution and comparing coefficients.

   • Substitute $x = 1$: The terms with $B$ and $(Cx+D)$ vanish because $(x-1)=0$. We get: $1 = A(2)(2) \implies 1 = 4A \implies A = \frac{1}{4}$.
   • Substitute $x = -1$: The $A$ and $(Cx+D)$ terms vanish because $(x+1)=0$. We get: $1 = B(-2)(2) \implies 1 = -4B \implies B = -\frac{1}{4}$.
   • Substitute $x = 0$: This gives a relation among all constants: $1 = A(1)(1) + B(-1)(1) + D(-1)(1) \implies 1 = A - B - D$. Plug $A = \frac{1}{4}$, $B = -\frac{1}{4}$: $1 = \frac{1}{4} - (-\frac{1}{4}) - D = \frac{1}{2} - D \implies D = -\frac{1}{2}$.
   • Compare coefficients of $x^3$ (or use another substitution, say $x=2$). The $x^3$ term on the right comes from $A(x)(x^2) = A x^3$, $B(x)(x^2) = B x^3$, and $(Cx)(x)(x) = C x^3$ (since $(Cx)(x-1)(x+1) = Cx(x^2-1)$ gives $C x^3$). So coefficient of $x^3$ is $A + B + C$. On the left, coefficient of $x^3$ is $0$. Thus: $A + B + C = 0 \implies \frac{1}{4} - \frac{1}{4} + C = 0 \implies C = 0$.

   So we have $A = \frac{1}{4}$, $B = -\frac{1}{4}$, $C = 0$, $D = -\frac{1}{2}$. …

Method: Decompose a Difference of Even Powers

Use this for denominators like $x^4-1$ that factor into linear factors plus an irreducible quadratic.

Steps

Step 1: Factor completely.

Apply difference of squares repeatedly:

The factor $x^2+1$ has no real roots, so it stays.

Step 2: Build the template — constants over linears, a linear numerator over the quadratic.

Step 3: Solve. …

Common Mistakes

Mistake 1: Trying to factor $x^2+1$ into real linear factors.

Why it's wrong: $x^2+1$ has no real roots, so it is irreducible over the reals and must stay as a quadratic factor. Correct approach: factor only to $(x-1)(x+1)(x^2+1)$.

Mistake 2: Putting a constant instead of $Cx+D$ over $x^2+1$. …