Problem 39 - Thomas Calculus

Question

Use a substitution to change the integral into one you can find in the table. Then evaluate the integral. \(\int \frac{x^{3}+x+1}{\left(x^{2}+1\right)^{2}} d x\)

Step-by-Step Solution

Answer

Use substitution \( u = x^2 + 1 \), rewrite integral, and evaluate using integration tables.

Step 1: Identify substitution

For the integral \( \int \frac{x^{3}+x+1}{(x^{2}+1)^{2}} dx \), we notice the presence of \(x^2 + 1\) in the denominator. Let's use the substitution \( u = x^2 + 1 \). This simplifies our expression.

Step 2: Differentiate and express dx in terms of du

Differentiate \( u \) with respect to \( x \) to find the new differential: \( du = 2x \ dx \). Solving for \( dx \), we have \( dx = \frac{du}{2x} \).

Step 3: Rewrite the integral in terms of u

Substitute \( u = x^2 + 1 \) and \( dx = \frac{du}{2x} \) into the integral: \[ \int \frac{x^{3} + x + 1}{(u)^{2}} \cdot \frac{du}{2x} = \frac{1}{2} \int \frac{x^3 + x + 1}{x(u)^2} \, du. \] We need to express everything in terms of \( u \). Since \( x^2 = u - 1 \), then \( x = \sqrt{u - 1} \). Substitute \( x\) back.

Step 4: Simplify and Break Integrals

Using \( x = \sqrt{u - 1} \), rewrite the numerator in terms of \( u \). Substitute \( x^3 = (\sqrt{u - 1})^3 \) and \( x = \sqrt{u - 1} \). Then break down the integral into simpler parts.

Step 5: Evaluate individual integrals

Split the integral into parts using the substitution and integration tables to find simpler integrals. Evaluate each part using integration techniques or a table, if needed.

Step 6: Add constant of integration

Once you integrate all parts, sum them up and add a constant of integration \( C \).

Step 7: Final Integration Result

Use the evaluations from the integration table and rewrite your result back in terms of \( x \) to provide the final answer.

Key Concepts

Integration Techniques

In calculus, integration is a fundamental concept that allows us to find areas under curves and solve various mathematical problems. There are several techniques available to tackle integrals successfully. Each method fits specific types of integrals and can simplify the process.

Some common techniques include:

• Substitution: Typically used when the integral contains a function and its derivative.
• Integration by Parts: Handy for products of functions, based on the product rule for differentiation.
• Partial Fraction Decomposition: Useful for breaking down functions into simpler fractions.
• Trigonometric Identities: Employs trigonometric functions and their identities to simplify the integration process.

It's crucial to analyze the given integral and choose the appropriate technique. This ensures simplification and accurate evaluation.

Substitution Method

The substitution method is a vital integration technique that simplifies integrals by changing variables. This method is particularly beneficial when dealing with composite functions. By introducing a new variable, we can transform a difficult integral into an easier one.

To apply substitution effectively:

• Identify a part of the integral that can be replaced by a new variable, usually something within a parenthesis like \(x^2 + 1\).
• Differentiate the chosen expression to get the corresponding \(du\).
• Replace \(dx\) using the differential equation \((dx = \frac{du}{2x})\).
• Rewrite the integral completely in terms of \('u'\) and solve.

This method not only makes calculations easier but also helps in revealing the integral function's underlying structure.

Definite and Indefinite Integrals

Integrals can be classified into definite and indefinite integrals. Understanding their differences is essential for correctly solving problems using integration techniques.

• Indefinite integrals represent the family of functions whose derivative is the integrand. It includes a constant of integration, \(C\), symbolizing all possible vertical shifts of the function. For instance, \(\int f(x)\, dx = F(x) + C\).
• Definite integrals provide a numerical value representing the area under the curve between two points, \(['a', 'b']\). It is expressed as \(\int_a^b f(x)\, dx\), emphasizing the integral as a function bounded by specific limits.

Indefinite integrals focus on finding the function, while definite integrals seek the precise area, aiding various applications such as physics and engineering.

Integration by Parts

Integration by parts is a technique useful for evaluating the integral of a product of functions, commonly derived from the product rule of differentiation.To perform integration by parts, follow this rule:\[\int u \, dv = uv - \int v \, du\]

• Identify parts of the integral: choose \(u\) and \(dv\), making sure that differentiating \(u\) is simpler, and integrating \(dv\) is feasible.
• Differentiate \(u\) to find \(du\), and integrate \(dv\) to receive \(v\).
• Substitute into the integration by parts formula.
• Solve the resulting integral, and don't forget to add the constant of integration when dealing with indefinite integrals.

Using integration by parts strategically helps tackle integrals that feature logs, inverse trigonometric functions, and polynomial-exponential combinations.