Problem 13 - The Calculus with Analytic Geometry

Question

\(\int \sqrt{1+\frac{1}{3 x}} \frac{d x}{x^{2}}\)

Step-by-Step Solution

Answer

\( -2 \left( 1 + \frac{1}{3x} \right)^{3/2} + C \).

Step 1 - Substitute Variable

Let us use substitution to simplify the integral. Let \( u = 1 + \frac{1}{3x} \). Then, calculate the differential \( du \). To do this, differentiate both sides of \( u = 1 + \frac{1}{3x} \). This gives \( du = -\frac{1}{3x^2} dx \). Thus, we have \( dx = -3x^2 du \).

Step 2 - Rewrite the Integral

Given \( x \) in terms of \( u \), rewrite the integral in terms of \( u \). Note that since \( u = 1 + \frac{1}{3x} \), we have \( x = \frac{1}{3(u-1)} \). Substituting these into the integral: \( \int \sqrt{u} \frac{du}{x^2} \). Substituting \( dx = -3x^2 du \) into the integral gives us \[ -3 \int \sqrt{u} du \].

Step 3 - Evaluate the Simplified Integral

Integrate \( -3 \int \sqrt{u} du \). The integral of \( \sqrt{u} \) is \( \frac{2}{3} u^{3/2} \). So, the integral becomes \( -3 \times \frac{2}{3} u^{3/2} = -2u^{3/2} \).

Step 4 - Substitute Back in Terms of x

Finally, substitute back \( u = 1 + \frac{1}{3x} \) to express the result in terms of \( x \). Thus, \[ -2 \left( 1 + \frac{1}{3x} \right)^{3/2} + C \], where \( C \) is the constant of integration.

Key Concepts

integration by substitution

Integration by substitution is a powerful method in integral calculus. It allows us to simplify complex integrals by changing variables. We often use substitution when the integrand contains a function and its derivative together.

Consider the integral \(\frac{dx}{x^2}\frac{1}{x^{2}}\). The first step is to choose a substitution that simplifies the integrand. In this example, we let \(u = 1 + \frac{1}{3x}\). Next, we find \(du\) by differentiating \(u\). This gives us \(du = -\frac{1}{3x^2} dx\). With the substitution, we replace \(dx\) resulting in \(dx = -3x^2 du\).

Transforming the integral into the new variable \(u\) often simplifies the problem. After integrating, remember to transform back to the original variable to get the final answer.

integral calculus

Integral calculus is one of the main branches of calculus. It deals with finding the integral of functions. Integration helps in determining areas under curves, volumes, and solving differential equations.

In practical terms, integral calculus is the process of finding the antiderivative of a function. For the given problem, our initial integral was \(\frac{dx}{x^2}\), which we simplified using substitution.

After substitution, we obtained the simpler integral \(-3 \(\frac{1}{3x}\) \sqrt{u} du\). Calculating the integral, we end up with \(\frac{d^{3}}\), which represents the antiderivative. Finally, we substitute back to the original variable to get the answer in terms of \( x \). This helps to understand the different techniques involved in integral calculus.

differentiation

Differentiation is essentially the opposite of integration. It involves finding the derivative of a function. While integration sums up areas, differentiation calculates rates of change.

To solve our integral using substitution, we needed to find \(du\). This involved differentiating the substitution function \(u = 1 + \frac{1}{3x}\). By differentiating, we found \(du = -\frac{1}{3x^2} dx\). This step is crucial as it allows us to replace \(dx\) and simplify the integral.

Differentiation is a foundational concept in calculus that not only assists in solving integrals through substitution but also finds applications in physics, engineering, and economics. Understanding how to differentiate accurately is key to mastering integral calculus.