Problem 62 - Applied Calculus

Question

Find each integral. [Hint: Separate each integral into two integrals, using the fact that the numerator is a sum or difference, and find the two integrals by two different formulas. \(\int \frac{x+1}{\sqrt{x^{2}+1}} d x\) [Hint: After separating into two integrals, find one by a formula and the other by a substitution.

Step-by-Step Solution

Answer

\( \int \frac{x+1}{\sqrt{x^2+1}} \, dx = \sqrt{x^2 + 1} + \ln |x + \sqrt{x^2+1}| + C \).

Step 1: Separate the Integral

The given integral is \( \int \frac{x+1}{\sqrt{x^{2}+1}} \, dx \). We can separate this integral into two different integrals by splitting the numerator:\[ \int \frac{x+1}{\sqrt{x^{2}+1}} \, dx = \int \frac{x}{\sqrt{x^{2}+1}} \, dx + \int \frac{1}{\sqrt{x^{2}+1}} \, dx \].

Step 2: Solve the First Integral

Let's solve \( \int \frac{x}{\sqrt{x^{2}+1}} \, dx \) using substitution. Set \( u = x^2 + 1 \), then \( du = 2x \, dx \) or \( \frac{1}{2} du = x \, dx \).Thus, the integral becomes:\[ \int \frac{x}{\sqrt{x^{2}+1}} \, dx = \frac{1}{2} \int \frac{1}{\sqrt{u}} \, du \].This integral evaluates to:\[ \frac{1}{2} \cdot 2\sqrt{u} + C = \sqrt{x^2 + 1} + C \].

Step 3: Solve the Second Integral

Now solve \( \int \frac{1}{\sqrt{x^{2}+1}} \, dx \). This integral is of the form \( \int \frac{1}{\sqrt{a^{2} + x^{2}}} \, dx \) which has the standard result:\[ \ln | x + \sqrt{x^2 + a^2} | + C \].In this case, \( a = 1 \), so the integral evaluates to:\[ \ln | x + \sqrt{x^2 + 1} | + C \].

Step 4: Combine Results

Combining both integrals, we have:\[ \int \frac{x+1}{\sqrt{x^{2}+1}} \, dx = \sqrt{x^2 + 1} + \ln | x + \sqrt{x^2 + 1} | + C \].This is the solution to the original integral, where \( C \) represents the constant of integration.

Key Concepts

Integration Techniques

Integration is a central concept in calculus that involves finding a function that produces a given derivative. There are different techniques that help in evaluating integrals, especially when they are not straightforward. Some common techniques include:

• Substitution Method: Makes integration simpler by transforming a difficult integral into a more manageable one.
• Integration by Parts: Useful when integrals are a product of functions.
• Partial Fraction Decomposition: Breaks down a complex rational function into simpler fractions.

In integral calculus, selecting the right technique is crucial for simplifying and accurately evaluating integrals.
In real-world problems and some exercises, a combination of techniques might be required for a solution.

Substitution Method

The substitution method is a powerful tool in integration, similar to the change of variables in algebra. It simplifies an integral by changing the variable of integration to make the integral more straightforward.
For example, in the exercise, we use substitution to solve the integral \( \int \frac{x}{\sqrt{x^{2}+1}} \, dx \). Here, we set \( u = x^2 + 1 \), which simplifies the integral to \( \frac{1}{2} \int \frac{1}{\sqrt{u}} \, du \).

• First, identify a substitution. Look for a function inside the integral that becomes simpler when differentiated. This new variable, \( u \), replaces the original variable.
• Adjust the differential accordingly. In our example, \( du = 2x \, dx \), so \( x \, dx = \frac{1}{2} du \).
• Integrate with respect to \( u \) and then substitute back the original variable to complete the process.

Substitution is effective when a component of the integrand is its own derivative or when you can easily transform the integral into a basic form.

Definite Integral

A definite integral computes the net area under a curve within a specific interval. However, in this exercise, we deal with indefinite integrals, where there are no limits of integration Ppresent.
For insight, definite integrals provide valuable tools for solving real-world problems involving accumulated quantities. Examples include:

• Finding the total distance traveled over a time span when velocity varies.
• Calculating the total amount of a substance that has flowed past a point over time.
• Determining areas between curves and axes.

When calculating a definite integral, the process involves using the fundamental theorem of calculus to find the difference in the function's antiderivatives at the bounds of the interval.

Indefinite Integral

Indefinite integrals represent a family of functions with an unspecified constant, known as the constant of integration, \( C \).
This constant is crucial, as indefinite integrals account for all possible antiderivatives a function may have.
In our example, after evaluating the separated integrals, we combine them and add \( C \) to ensure all potential solutions are included. The general result reflects all the functions whose derivative will give the integrand.

• They can describe the general solution to a differential equation.
• Useful in understanding translation within space due to the added constant \( C \).
• Essential for solving physical problems where initial conditions are unknown.

Understanding indefinite integrals is foundational to grasping the complete nature of solutions in integral calculus.