Question

Identify \(u\) and \(d u / d x\) for the integral \(\int u^{n}(d u / d x) d x\). $$ \int \frac{1}{(1+2 x)^{2}}(2) d x $$

Short answer

The expressions obtained for this problem are \(u = 1 + 2x\) and \(du/dx = 2\).

Step-by-step solution

Identify 'u'

We can identify \(u\) by looking for a function inside the integral that will simplify the integral if taken as \(u\). In this integral, \(u\) is taken as \(1 + 2x\) which makes the integral simpler.

Identify 'du/dx'

The \(d u / d x\) is also a term in integral. In our fundamental formula for integration i.e. \(\int u^{n}(d u / d x) d x\), the \((d u / d x)\) is general notation for deriving \(u\) with respect to \(x\). So here, \(d u / d x\) is identified as the derivative of \(u\) with respect to \(x\), this gives us \(\frac{d u}{d x} = 2\). So, \(d u = 2 dx\)

Key concepts

Indefinite Integral

An indefinite integral can be thought of as the reverse operation of taking a derivative. It is used to find the function that originally was differentiated to produce the given derivative. This process is also known as anti-differentiation. An indefinite integral is usually written in the form \(\int f(x)\,dx\) and represents a family of functions, F(x), whose derivative is f(x). It's called 'indefinite' because it contains an arbitrary constant, C, since integrating a function and then differentiating it will leave the constant out of the result.
For example, if we have \(f(x) = x^2\), then an indefinite integral of f(x) is \(F(x) = \frac{1}{3}x^3 + C\), where C can be any constant. The presence of the arbitrary constant is due to the fact that any number of constants can be added to a function without changing its derivative.

U-Substitution

U-substitution is a technique used to simplify the integration process by substituting a part of the integral with a new variable, typically \(u\), which makes the integral easier to evaluate. This method is particularly useful when an integral contains a function and its derivative. Here's how it works:

1. Choose a part of the integral as \(u\) and differentiate it with respect to \(x\) to find \(\frac{du}{dx}\).
2. Rewrite the integral in terms of \(u\) by substituting \(dx\) with \(\frac{du}{\frac{du}{dx}}\).
3. Evaluate the integral using standard formulas or methods, treating it as an integral in terms of \(u\) alone.
4. Finally, substitute back the original \(x\)-terms to express the antiderivative in terms of the original variable \(x\).

In the given exercise, we took \(u = 1 + 2x\) and found \(du/dx = 2\), which allowed us to reduce the integral to a simpler form.

Derivative

The derivative represents the rate at which a function's value is changing at any given point and is a fundamental concept in calculus. It is denoted by \(\frac{dy}{dx}\) or \(f'(x)\), among other notations, where \(y = f(x)\). The process of finding a function's derivative is known as differentiation.
In the context of u-substitution, identifying the derivative of \(u\) with respect to \(x\) is crucial because it allows us to link \(dx\) to \(du\), which in turn simplifies the integral. In our exercise, we identified that \(\frac{du}{dx} = 2\), which means for a small change in \(x\), \(u\) changes at twice that rate. This relationship helps us transform the original integral into a form that is more straightforward to integrate by substituting \(dx\) with \(du/2\).

Antiderivative

The antiderivative, often known as the primitive, is the inverse operation to differentiation. It is the process of finding a function \(F(x)\) whose derivative is the given function f(x). Essentially, if \(F'(x) = f(x)\), then \(F(x)\) is an antiderivative of \(f(x)\).
When we perform integration, we are essentially calculating an antiderivative. Since the derivative of a constant is zero, there are infinitely many antiderivatives for any function f(x), each differing by a constant. Thus, when we find an antiderivative, we always add a constant \(C\) to represent the general solution to the integral. In practice, finding the antiderivative involves recognizing patterns and applying integral formulas, often aided by techniques like u-substitution, as demonstrated in the exercise with \(u = 1 + 2x\).