Question

Determine the following indefinite integrals. \(\int \frac{d x}{x \sqrt{1+x^{4}}}\)

Short answer

Question: Find the indefinite integral of the function $$\frac{1}{x\sqrt{1+x^4}}$$ with respect to x. Answer: The indefinite integral of the function $$\frac{1}{x\sqrt{1+x^4}}$$ with respect to x is $$\frac{1}{2}(\arcsin(x^2) + C)$$, where C is the constant of integration.

Step-by-step solution

Define the substitution (u-substitution)

We will first define a substitution that will simplify our integral. Let's define a variable \(u\) such that: $$u = x^2$$

Find the derivative of the substitution

Now we need to find the differential of our substitution \(u\). $$\frac{du}{dx} = 2x$$

Replace dx and the terms

Solving for \(dx\), we get $$dx = \frac{du}{2x}$$ Replace \(x^2\) with \(u\) in our integral and replace \(dx\) with the expression we just found: $$\int \frac{1}{x\sqrt{1+x^4}}dx = \int \frac{1}{x\sqrt{1+u^2}}\cdot\frac{du}{2x}$$

Simplify the integral

We can now simplify the integral: $$\int \frac{1}{x\sqrt{1+x^4}}dx = \frac{1}{2}\int \frac{1}{\sqrt{1+u^2}}du$$

Recognize the standard integral and integrate

We can now recognize the standard integral form $$\int \frac{1}{\sqrt{1+u^2}}du$$ which is the arcus sine integral. The result of this integral is: $$\frac{1}{2}(\arcsin(u) + C)$$

Substitute back the original variable

Now we can substitute back the expression for \(u\) in terms of \(x\): $$\frac{1}{2}(\arcsin(x^2) + C)$$ So the solution to the indefinite integral is: $$\int\frac{1}{x\sqrt{1+x^4}}dx = \frac{1}{2}(\arcsin(x^2) + C)$$

Key concepts

Substitution Method

The Substitution Method is a powerful tool in calculus for finding indefinite integrals. It's like solving a puzzle by replacing a complex part with something simpler.
Imagine you have a difficult integral, such as \(\int \frac{d x}{x \sqrt{1+x^{4}}}\). To simplify it, you can substitute a part of the integral with a new variable. In this case, we choose \(u = x^2\).
This transforms the problem into a simpler form by changing variables. Following the substitution, you find the derivative and express \(dx\) in terms of \(du\): \(\frac{du}{dx} = 2x\) leads to \(dx = \frac{du}{2x}\).
By substituting \(x^2\) with \(u\) and \(dx\) with our expression in the integral, you create an easier integral to solve. This substitution turns a challenging integral into a recognizable and simpler one.

• Substitute challenging parts with a new variable \(u\)
• Find the derivative to replace \(dx\)
• Simplify and solve the transformed integral

This method is super helpful when encountering integrals with compositions of functions or when known integral forms can be exploited.

Integral Calculus

Integral Calculus is all about finding the area under curves or, more generally, the antiderivatives of functions.
Indefinite integrals, like \(\int \frac{d x}{x \sqrt{1+x^{4}}}\), represent a family of functions whose derivative gives the original function under the integral sign.
Indefinite integrals include a constant, \(C\), which accounts for the fact that antiderivatives are not unique.
Calculating indefinite integrals often involves applying techniques such as substitution or integration by parts to reach an integral with a known formula.
The computed integral \(\int \frac{1}{x\sqrt{1+x^4}}dx = \frac{1}{2}(\arcsin(x^2) + C)\) showcases how a definitive substitution and simplification can reveal a standard integral form. This makes integration not only feasible but manageable.

• Seek the antiderivative of given functions
• Use constants \(C\) for generality
• Apply strategic methods to simplify integrals

Integral calculus empowers us to solve problems involving accumulation of quantities, such as areas and other applications involving rates of change.

Trigonometric Functions

Trigonometric Functions frequently appear in calculus, both as parts of the function being integrated and in the results.
In the integral \(\int \frac{1}{x\sqrt{1+x^4}}dx\), the result \(\frac{1}{2}(\arcsin(x^2) + C)\) leverages the arc sine function, one of the inverse trigonometric functions.
Inverse trigonometric functions like arcsine (\(\arcsin\)) are essential as they often form the solutions in integration.
Recognizing these relationships allows us to identify and manipulate integrals into standard forms, helping in solving them efficiently.

• Involve functions like sine, cosine, and their inverses
• Used in integration to transform and simplify expressions
• Key in identifying standard forms for solutions

The arc sine function, in particular, connects the integral with geometric interpretations, like finding angles whose sine is a given value. Such connections are key in solving integrals involving trigonometric expressions.