Question

Evaluate each integral. $$\int \frac{d x}{x \sqrt{1+x^{4}}}$$

Short answer

Question: Determine the integral of the following function: $$\int \frac{d x}{x \sqrt{1+x^{4}}}$$ Answer: $$\int \frac{d x}{x \sqrt{1+x^{4}}} = \frac{1}{2} \sinh^{-1}(x^2) + C$$

Step-by-step solution

Choose the substitution

Let us choose the substitution \(u = x^2\). We then differentiate \(u\) with respect to \(x\) to find \(du\), which will help us to rewrite the integral in terms of \(u\): $$du = \frac{d u}{d x} d x = 2 x d x$$ Now, we can solve for \(dx\) as follows: $$dx = \frac{1}{2 x} du$$

Rewrite the integral in terms of \(u\)

Substitute the expressions for \(u\) and \(dx\) into the given integral: $$\int \frac{d x}{x \sqrt{1+x^{4}}} = \int \frac{1}{2 x} \frac{d u}{\sqrt{1+(x^{2})^{2}}}$$ Now, substitute \(u = x^2\): $$\int \frac{1}{2 x} \frac{d u}{\sqrt{1+u^2}}$$

Cancel out the \(x\) term and integrate with respect to \(u\)

Notice that we can cancel out the \(x\) term in the integral: $$\frac{1}{2} \int \frac{d u}{\sqrt{1+u^2}}$$ Now we can integrate this expression with respect to \(u\). The integral can be evaluated using a well-known result for the inverse hyperbolic sine: $$\frac{1}{2} \int \frac{d u}{\sqrt{1+u^2}} = \frac{1}{2} \sinh^{-1}(u) + C$$ where \(C\) is the integration constant.

Substitute back the value of \(x\)

Finally, we need to substitute back the value of \(x\) we used earlier for the substitution. Recall that \(u = x^2\): $$\frac{1}{2} \sinh^{-1}(u) + C = \frac{1}{2} \sinh^{-1}(x^2) + C$$ Thus, the final result of the integral is: $$\int \frac{d x}{x \sqrt{1+x^{4}}} = \frac{1}{2} \sinh^{-1}(x^2) + C$$

Key concepts

Integration by Substitution

Integration by substitution is a powerful tool used in calculus to simplify the process of finding indefinite integrals. This technique, also known as u-substitution, involves changing the variable in an integral to make it easier to evaluate.

The first step is to choose an appropriate substitution, which is usually a function inside the integral that, when differentiated, appears elsewhere in the integral. For example, in the given exercise, we let the substitution be \( u = x^2 \). The next step is to calculate the differential \( du \) and express \( dx \) in terms of \( du \), which allows us to rewrite the original integral in terms of \( u \).

Applying this method transforms the integral into a simpler form, which is easier to evaluate. Once the integral is found in terms of \( u \), we switch back, replacing \( u \) with the original variable to complete the process. A crucial part of performing integration by substitution is keeping track of both the new variable and its differential to avoid mistakes and ensure the transformation is correct.

Inverse Hyperbolic Functions

Inverse hyperbolic functions are reciprocal functions to the standard hyperbolic functions and are used extensively in calculus, especially in indefinite integral problems where the normal trigonometric substitutions fail. These functions include hyperbolic sine \( \sinh(x) \), hyperbolic cosine \( \cosh(x) \), and hyperbolic tangent \( \tanh(x) \), among others.

The inverse hyperbolic sine, \( \sinh^{-1}(x) \), which appeared in our solution, is equivalent to the natural logarithm expression \( \ln(x + \sqrt{x^2 + 1}) \). This identity simplifies the process of integrating terms of the form \( \frac{du}{\sqrt{1+u^2}} \), because the integral of this expression is directly \( \sinh^{-1}(u) + C \), where \( C \) is the constant of integration. Understanding the inverse hyperbolic functions and their properties allows students to work through complex integral calculus problems more efficiently.

Integral Calculus Techniques

Integral calculus techniques are a collection of methods used to evaluate integrals, the inverse process of differentiation. These methods include integration by substitution, integration by parts, partial fraction decomposition, trigonometric substitution, and miscellaneous approaches like the use of tables or computer algebra systems.

Each method has its own set of rules and applications, suited for different kinds of integrand functions. For instance, integration by parts is useful when the integrand is a product of two functions, while partial fraction decomposition is applied when dealing with rational functions.

Students are often encouraged to familiarize themselves with various techniques, as sometimes multiple methods can be applied to the same problem. This flexibility in approach helps build problem-solving skills in calculus. Successful integration often relies on recognizing which method to apply and combining these techniques effectively, as seen in the original exercise where substitution and knowledge of inverse hyperbolic functions led to the solution.