Q.Evaluate: $\int \dfrac{x}{x^4-1}\,dx$

Concept: Partial Fraction Decomposition — factor the denominator and split the integrand into simpler fractions.

Step 1: Factor the denominator.

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

Step 2: Set up the partial fractions.

Multiplying through and comparing coefficients gives $A = \frac14$, $B = \frac14$, $C = -\frac12$, $D = 0$.

Step 3: Integrate term by term.

$\int \frac{1/4}{x-1}\,dx + \int \frac{1/4}{x+1}\,dx - \int \frac{x/2}{x^2+1}\,dx$ …

The key insight here is that the numerator $x$ is almost the derivative of $x^2$, and the denominator $x^4-1$ is a difference of squares in $x^2$. This makes a substitution the natural first move — it transforms the integral into a rational function of $u$, which we can then decompose using partial fractions.

Let’s walk through it.

1. Substitute $u = x^2$. Then $du = 2x\,dx$, so $x\,dx = \frac{du}{2}$. The integral becomes

The denominator $u^2-1$ factors as $(u-1)(u+1)$, so we’re now integrating a proper rational function.

2. Set up the partial fraction decomposition. We want constants $A$ and $B$ such that

Multiply through by $u^2-1$:

This identity must hold for all $u$.

3. Solve for $A$ and $B$. Two clean ways:
   • Plug convenient values. Let $u = 1$: then $1 = A(2) + B(0) \Rightarrow A = \frac12$. Let $u = -1$: then $1 = A(0) + B(-2) \Rightarrow B = -\frac12$.
   • Equate coefficients. Expanding: $1 = (A+B)u + (A-B)$. Comparing coefficients gives $A+B=0$ and $A-B=1$, which yields the same result. Either way,

4. Integrate term by term. …

Method: Substituting $t=x^2$ for even-symmetric rational integrands

Use this when a rational integrand can be written as $x$ times a function of $x^2$ — e.g. $\dfrac{x}{x^4-1}$. The substitution $t=x^2$ turns it into a lower-degree rational integral.

Steps

Step 1: Spot the $x\cdot(\text{function of }x^2)$ pattern.

Here $\dfrac{x}{x^4-1}=\dfrac{x}{(x^2)^2-1}$, with a single $x$ available for the differential.

Step 2: Substitute $t=x^2$, $dt=2x\,dx$.

Step 3: Integrate the standard $\dfrac{1}{t^2-1}$ form. …

Common Mistakes

Mistake 1: Jumping to full partial fractions in $x$.

Why it's wrong: decomposing $\tfrac{x}{x^4-1}$ over four linear/quadratic factors is far more work than $t=x^2$. Correct approach: use the $x\,dx=\tfrac12dt$ substitution first.

Mistake 2: Wrong standard form for $\tfrac{1}{t^2-1}$.

Why it's wrong: $\int\tfrac{dt}{t^2-1}=\tfrac12\log|\tfrac{t-1}{t+1}|$, not $\tan^{-1}$ (that is for $t^2+1$). Correct approach: for $t^2-1$ use the log form. …