Problem 265 - Calculus Volume 1

Question

In the following exercises, find the antiderivative using the indicated substitution. $$ \int \frac{x}{\sqrt{x^{2}+1}} d x ; u=x^{2}+1 $$

Step-by-Step Solution

Answer

The antiderivative is \( \sqrt{x^2 + 1} + C \).

Step 1: Identify the substitution

We are given the substitution \( u = x^2 + 1 \). This means the expression under the square root \( \sqrt{x^2 + 1} \) will be replaced by \( \sqrt{u} \).

Step 2: Find the derivative of substitution

Differentiate \( u \) with respect to \( x \): \( \frac{du}{dx} = 2x \). This implies \( du = 2x \, dx \) or \( \frac{1}{2}du = x \, dx \).

Step 3: Rewrite the integral using substitution

Substitute \( u = x^2 + 1 \) and \( x \, dx = \frac{1}{2} du \) into the integral: \( \int \frac{x}{\sqrt{x^2 + 1}} \, dx = \int \frac{1}{\sqrt{u}} \cdot \frac{1}{2} du \).

Step 4: Simplify the integral

The integral simplifies to \( \frac{1}{2} \int u^{-1/2} \, du \).

Step 5: Integrate using the power rule

Apply the power rule for integration: \( \int u^{n} \, du = \frac{u^{n+1}}{n+1} + C \). So, \( \int u^{-1/2} \, du = 2u^{1/2} + C \). Multiply by \( \frac{1}{2} \): \( \frac{1}{2} \cdot 2u^{1/2} = u^{1/2} + C \).

Step 6: Back-substitute to original variable

Replace \( u \) with \( x^2 + 1 \) to obtain the antiderivative in terms of \( x \): \( \sqrt{x^2 + 1} + C \).

Key Concepts

Substitution Method

In calculus, the substitution method is a helpful technique to simplify integration problems. It's like changing variables in an algebra problem to make things easier to solve.
First, you start by identifying a part of the integral that can be substituted with a new variable, typically denoted as "u". This new variable should simplify the integrand, making the integration process more straightforward.
For our exercise, we identified that the expression under the square root, \(x^2 + 1\), is a good candidate for substitution. So, we let \(u = x^2 + 1\). Then, we find the derivative of \(u\) with respect to \(x\), which is \(2x\). This helps in substituting \(dx\) in the integral with \(du\). After substituting, the integral becomes much simpler to evaluate.

• Make a substitution \(u = x^2 + 1\).
• Find \(du = 2x \, dx\).
• Replace \(x \, dx\) with \(\frac{1}{2} du\).

Using the substitution method requires practice, but it often turns a complex problem into a manageable one.

Power Rule

The power rule is a fundamental tool in calculus for finding antiderivatives and derivatives. For integration, the power rule states that the integral of \(u^n\) with respect to \(u\) is \(\frac{u^{n+1}}{n+1} + C\), where \(n eq -1\). This rule helps simplify integration tasks involving simple power terms.
Applying the power rule to our exercise, we had to integrate \(u^{-1/2}\) after substitution. Here's how it works:
1. Identify the exponent of \(u\), which is \(-1/2\).
2. Increase the exponent by one to get \(1/2\).
3. Divide by the new exponent, leading to the expression \(2u^{1/2}\).
4. Don't forget to add the constant of integration, \(C\).
Finally, we multiplied the result by the constant factor \(\frac{1}{2}\) from our substitution step. The power rule is simple but powerful for tackling polynomial expressions even in their root forms.

Calculus

Calculus is the branch of mathematics that studies how things change. It provides tools for understanding the behavior of functions and helps solve problems involving rates of change and areas under curves.
At its core, calculus is divided into two main parts: differentiation and integration. Differentiation deals with finding the rate of change, while integration focuses on finding the total size or value from the rate of change.
In our context, we are dealing with integral calculus, which finds the antiderivatives or the original functions given their derivatives. This exercise particularly teaches us to simplify complex integrals using clever substitutions and techniques like the power rule.

• Calculus helps understand the physical world and systems.
• It comprises differentiation and integration.
• Integral calculus is used to find areas and volumes.

Calculus can seem complex at first, but it's an incredibly valuable tool for scientific and engineering applications.

Integration

Integration is a process in calculus used to find functions whose derivative is the given function, essentially reversing differentiation. Integrating a function helps us calculate areas, volumes, and other quantities when given their rates of change.
For instance, finding the antiderivative in our exercise means determining a function that, when differentiated, returns the original integrand \(\frac{x}{\sqrt{x^2+1}}\).
With our substitution \(u = x^2+1\), the integration becomes feasible:

• We transformed the original integral into \(\frac{1}{2} \int u^{-1/2} \, du\).
• Using the power rule, we integrated to find \(u^{1/2}\).
• Back Substitution returned the solution in terms of \(x\), \(\sqrt{x^2+1} + C\).

Integration is not just about solving equations; it's a gateway to understanding the continuous accumulation of quantities, underpinning many real-world phenomena.