If \(\int\left[\left(x^{4}+1\right) /\left(x^{6}+1\right)\right] d x=\tan ^{-1} x+P \tan ^{-1} x^{3}+c\) then \(P=\) (a) 3 (b) \((1 / 3)\) (c) \(-(1 / 3)\) (d) \(-3\)

Understanding indefinite integrals is a fundamental step in grasping calculus. An indefinite integral, also known as an antiderivative, represents a family of functions whose derivative is the given function.

For instance, when integrating a function such as \( f(x) \), the indefinite integral is represented as \( \int f(x) \, dx \), which essentially gives us all possible functions \( F(x) \), such that \( F'(x) = f(x) \). It is important to note that the indefinite integral includes a constant of integration, \( c \), since the derivative of a constant is zero. This constant represents an infinite number of functions that differ from each other by a constant value.

For example, given \( \int (x^4 + 1)/(x^6 + 1) \, dx \), we aim to find the function \( F(x) \), for which \( F'(x) = (x^4 + 1)/(x^6 + 1) \). The indefinite integral indicates a family of solutions, adding the constant \( c \) for completeness.

The technique known as integration by parts derives from the product rule for differentiation and is used to integrate the product of two functions.

The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \), where \( u \), and \( dv \), are parts of the original integral that we choose strategically. The goal is to simplify the integration process by breaking it down into more manageable parts.

When choosing \( u \) and \( dv \) from our function, it’s beneficial to select \( u \) to be a function that is easy to differentiate, and \( dv \) to be a function easily integrable. However, this technique was not used in the original problem, but it's a helpful strategy in many integration problems where the integrand is a product of two different kinds of functions.

Integration involving inverse trigonometric functions can often appear in calculus problems. These functions, like \( \tan^{-1}(x) \), represent the reverse process of the usual trigonometric functions.

For example, the derivative of \( \tan^{-1}(x) \) is \( \frac{1}{1+x^2} \), and thus the integral becomes relevant when encountering a fraction that resembles the derivative form. In our exercise, integration involved inverse trigonometric functions, where \( \tan^{-1}(x) \) and \( \tan^{-1}(x^3) \) appeared as parts of the given integral function.

The ability to recognize when to apply the knowledge of inverse trigonometric integrals is crucial. Knowing the derivatives of inverse trigonometric functions helps one integrate these types of functions when they are present in an integrand.

The chain rule is essential for differentiating composite functions—functions made by combining two or more functions. It applies to situations where you need to find the derivative of a function within another function.

The chain rule states that if you have a composite function \( h(x) = f(g(x)) \), then the derivative of \( h \) with respect to \( x \) is \( h'(x) = f'(g(x)) \cdot g'(x) \).

In the given solution, the chain rule was used to differentiate \( P\tan^{-1} x^3 \), resulting in \( \frac{3Px^2}{1+x^6} \). This process is vital because it allows us to find the derivative that is needed to solve integration problems when inverse trigonometric functions and other composite functions are involved.