Problem 2 - Applied Calculus

Question

Find the integrals in problems. Check your answers by differentiation. $$ \int \frac{2 x}{x^{2}+1} d x $$

Step-by-Step Solution

Answer

The integral is \( \ln(x^2 + 1) + C \).

Step 1: Identify the Form of the Integral

The integral \( \int \frac{2x}{x^2+1} \, dx \) resembles a form that can be resolved using substitution. The denominator \( x^2 + 1 \) suggests using \( u = x^2 + 1 \) as a substitution.

Step 2: Perform the Substitution

Let \( u = x^2 + 1 \). Then, the derivative \( \frac{du}{dx} = 2x \), or \( du = 2x \, dx \). This substitution makes the integral \( \int \frac{1}{u} \, du \).

Step 3: Integrate with Respect to the New Variable

The integral \( \int \frac{1}{u} \, du \) is a standard form that equals \( \ln |u| + C \), where \( C \) is the constant of integration.

Step 4: Substitute Back the Original Variable

Replace \( u \) with \( x^2 + 1 \) to express the integral in terms of the original variable: \( \ln |x^2 + 1| + C \). Since \( x^2 + 1 \) is always positive, we can drop the absolute value: \( \ln(x^2 + 1) + C \).

Step 5: Differentiate to Check Answer

Differentiate \( \ln(x^2 + 1) + C \) with respect to \( x \). The derivative is \( \frac{1}{x^2 + 1} \cdot 2x = \frac{2x}{x^2 + 1} \), which matches the original integrand \( \frac{2x}{x^2 + 1} \). This confirms our solution is correct.

Key Concepts

Integration by Substitution

Integration by substitution is a lovely technique that simplifies the integration process by changing the variable of integration to make the integral easier to evaluate. The essential idea is to find a way to replace a complicated part of the integrand with a single variable, which then simplifies the entire problem.

Here's how it works in our problem:

• First, recognize the pattern or part of the function that can be substituted. In this case, we noticed that the denominator, \(x^2 + 1\), is a good candidate for substitution.
• Set \(u = x^2 + 1\). This substitution changes our variable and allows us to express the entire integral in terms of \(u\), rather than \(x\).
• Calculate \(du/ dx\), which is the derivative of \(u\). Here, we found that \(du = 2x \, dx\). This means our integral becomes \(\int \frac{1}{u} \, du\), a much simpler form!

With the substitution complete, the math becomes straightforward, allowing us to integrate more easily.

Definite Integral

In integral calculus, the term 'definite integral' might have appeared a bit confusing, but it's straightforward once you see how it works. Unlike indefinite integrals that have a constant of integration, definite integrals compute the net area under the curve from one point to another.

However, in this exercise, we focused on solving an indefinite integral using substitution. Here, we didn't apply the limits of integration, which would convert it to a definite integral. Nevertheless, the process for solving remains critically linked:

• The substitution technique is equally valuable in solving both indefinite and definite integrals.
• After substitution and integration, if there were limits, substitute back into the original variable to apply those limits for the final result.

So, understanding substitution as a technique opens a door to tackling definite integrals more efficiently once you get there!

Checking Integration Using Differentiation

A great habit in calculus is verifying your integration by differentiating your result. This practice ensures that you've carried out the process correctly, as differentiation acts like a magnifying glass to check every step of your integration.

Let's see how it's done with our solution from earlier:

• After performing the substitution and integration, we derived the expression \(\ln(x^2 + 1) + C\).
• To check, differentiate \(\ln(x^2 + 1)\) with respect to \(x\). Remember that the derivative of \(\ln(u)\) is \(\frac{1}{u} \frac{du}{dx}\).
• Here, this becomes \(\frac{1}{x^2 + 1} \cdot 2x\), which simplifies back to the original function \(\frac{2x}{x^2 + 1}\).

When differentiation returns you to the initial integrand, you can be confident that your integration was accurate. It's like checking the math puzzles you've solved, to make sure every piece fits into the right place!