Q.Find: $\displaystyle\int \dfrac{x^3 + 1}{x^3 - x}\, dx$

The key idea is to integrate a rational function by first simplifying it using polynomial long division, then applying partial fraction decomposition.

First, rewrite the integrand by performing polynomial long division or algebraic manipulation:

Next, factor the denominator of the fractional part and simplify:

Now, decompose $\frac{1}{x(x - 1)}$ into partial fractions:

$\frac{1}{x(x - 1)} = \frac{A}{x} + \frac{B}{x - 1} \implies 1 = A(x - 1) + Bx$ …

When faced with an integral of a rational function, the first step is always to compare the degrees of the numerator and the denominator. If the degree of the numerator is greater than or equal to the degree of the denominator, we must perform polynomial long division (or an equivalent algebraic manipulation) to rewrite the integrand as a polynomial plus a proper rational function (where the numerator's degree is strictly less than the denominator's degree). This makes the integral manageable, as polynomials are easy to integrate, and proper rational functions can be tackled using partial fraction decomposition.

In this problem, both the numerator ($x^3 + 1$) and the denominator ($x^3 - x$) have a degree of $3$. Therefore, we begin by simplifying the expression.

1. Perform Algebraic Manipulation (Polynomial Division): Since the degree of the numerator is equal to the degree of the denominator, we can rewrite the fraction by adding and subtracting terms in the numerator to match the denominator.

We can then split this into two terms:

Now, the original integral becomes:

The first part, $\int 1\, dx$, is simply $x + C_1$. We now focus on the second integral.

2. Factor the Denominator:

To apply partial fraction decomposition to $\dfrac{x + 1}{x^3 - x}$, we first need to factor the denominator completely.

Recognizing $x^2 - 1$ as a difference of squares, we factor it further:

So the rational term becomes $\dfrac{x + 1}{x(x - 1)(x + 1)}$.

3. Simplify and Perform Partial Fraction Decomposition:

Observe that there is a common factor of $(x + 1)$ in both the numerator and the denominator. For $x \neq -1$, we can simplify the expression:

This simplification is valid for integration because the integral is defined over intervals where the function is continuous, and a single point of discontinuity ($x=-1$) does not affect the value of the integral or the form of the antiderivative.

Now, we decompose $\dfrac{1}{x(x - 1)}$ into partial fractions. We assume the form:

To find $A$ and $B$, we multiply both sides by $x(x - 1)$:

We can find $A$ and $B$ by substituting convenient values for $x$:
*   Set $x = 0$:

*   Set $x = 1$: …