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Understanding how to simplify complex algebraic expressions is crucial in integral calculus. Polynomial long division is a technique used to divide a polynomial by another polynomial of equal or lower degree. Just like long division in arithmetic, this method helps break down intimidating integrals into more manageable terms.

For instance, consider the integral of a rational function, where the numerator's degree is higher than or equal to the denominator. In the given exercise, the function \( \frac{x^{3}}{x^{2}+1} \) initially seems challenging to integrate. However, by performing polynomial long division, we can rewrite this function as \( x - \frac{x}{x^{2} + 1} \), which is significantly easier to work with. This simplification allows us to address each part of the expression separately, turning a complicated integral into a sum or difference of simpler integrals.

A power function is of the form \( x^n \) where \( n \) is any real number. The process of integrating power functions is a fundamental concept in calculus and frequently appears in various applications. The integral of a power function \( x^n \) with respect to \( x \) is \( \frac{x^{n+1}}{n+1} + C \) as long as \( n \) is not equal to -1, where \( C \) is the constant of integration.

In our exercise, we apply this rule to the term \( \int x dx \) resulting from the polynomial long division. The integration proceeds smoothly, giving us \( \frac{x^2}{2} + C \) as the outcome for this part of the integral. Remember that understanding how to integrate power functions is crucial when dealing with polynomials, a common occurrence in integral calculus.

The derivative of the arctangent function or inverse tangent function, denoted as \( \arctan(x) \), plays an interesting role in integration. The function \( \arctan(x) \) differentiates to \( \frac{1}{x^2 + 1} \), making it particularly useful when encountering integrals involving expressions like \( \frac{x}{x^2 + 1} \).

In the exercise, the second part of the split integral after polynomial long division is \( -\int \frac{x}{x^2 + 1} dx \). Recognizing that the integrand resembles the derivative of \( \arctan(x) \) indicates that we can integrate it directly to obtain \( -\arctan(x) + C \), with \( C \) being the constant of integration. This connection between a function and its derivative is an example of how differential calculus intertwines with integral calculus to solve complex problems.