Problem 31 - Applied Calculus

Question

Find each indefinite integral by the substitution method or state that it cannot be found by our substitution formulas. $$ \int e^{x^{2}+2 x+5}(x+1) d x $$

Step-by-Step Solution

Answer

\(\frac{1}{2} e^{x^2 + 2x + 5} + C\)

Step 1: Identify the Inner Function

First, we need to identify a part of the integrand that can be represented as a function inside another function, suitable for substitution. Looking at the expression inside the exponential, we have \(x^2 + 2x + 5\). Let's set this inner function \(u = x^2 + 2x + 5\).

Step 2: Differentiate the Inner Function

To use substitution, we need \(du\), the differential of \(u\). Differentiate \(u = x^2 + 2x + 5\) with respect to \(x\): \[du = (2x + 2) \, dx.\]

Step 3: Match the Derivative with Part of the Integrand

Now, we check our integral \(\int e^{x^{2}+2x+5}(x+1) \, dx\) to see if it contains the derivative \(2x+2\) from Step 2. Rewrite \(2x+2\) as \(2(x + 1)\) to make it easier to compare. Notice that our integrand includes \((x+1)\).

Step 4: Adjust and Substitute

Rewrite \(2(x + 1)\) to match the differential needed for substitution:\(du = 2(x + 1) \, dx\).Thus, we need \(\frac{1}{2}du = (x+1) \, dx\).Substitute into the original integral:\[\int e^{u} \cdot \frac{1}{2} du.\]

Step 5: Integrate the Simplified Expression

Now, the expression \(\int \frac{1}{2} e^u \, du\) is a standard integral. Integrate:\[\frac{1}{2} \int e^u \, du = \frac{1}{2} e^u + C,\]where \(C\) is the constant of integration.

Step 6: Substitute Back to Original Variable

Finally, substitute back \(u = x^2 + 2x + 5\) into the expression:\[\frac{1}{2} e^{x^2 + 2x + 5} + C.\]

Key Concepts

Substitution Method

The substitution method is a powerful technique for solving integrals. It involves finding a part of the integrand, called the "inner function," and substituting it with a single variable. This simplification often transforms a complex integral into a more manageable form. In our example, the inner function was identified as \(u = x^2 + 2x + 5\).
Once the substitution is made, the goal is to replace every \(x\) in the original integral with expressions in terms of \(u\). This requires us to find \(du\), the differential of \(u\). We calculate \(du = (2x + 2) \, dx\).
If the expression \(du\) or a suitably scaled version of it matches part of the integrand, the substitution has successfully simplified the integral, making it ready for integration.

Integration by Parts

Integration by parts is another essential technique in calculus. It is particularly useful when dealing with products of functions, where one function is easily differentiable, and the other is easily integrable. The method is based on the product rule for differentiation and can be remembered by the formula:
\[\int u \, dv = uv - \int v \, du\]
This method wasn't needed in our example, but it's important to recognize when to use it. It is usually employed when a direct substitution does not simplify the problem, specifically when faced with an integral of a product of two different types of functions.

Integration Techniques

There are various integration techniques available to find indefinite integrals, such as substitution, integration by parts, partial fractions, and trigonometric substitution. Each method has its own set of suitable cases for application.

• **Substitution** is useful when the integral contains a function and its derivative.
• **Integration by Parts** helps solve integrals involving products of functions.
• **Trigonometric Substitution** is effective for integrals involving square roots of quadratic expressions.
• **Partial Fractions** are suitable for integrals of rational functions.

Choosing the right technique often depends on recognizing patterns within the integrand. It involves experience and practice to determine which technique will simplify the problem.

Calculus Integration

Calculus integration is a fundamental concept involving the process of finding antiderivatives. There are two main types: definite integrals, which calculate area under a curve, and indefinite integrals, which focus on finding antiderivatives.
Indefinite integration involves solving integrals without limits and includes a constant of integration \(C\). The techniques discussed are tools for tackling these indefinite integrals.
In practice, real-world problems often require transforming complex expressions into simpler forms using substitution or by selecting appropriate integration methods, much like solving puzzles. Mastery of integration is key in applying calculus to fields like physics, engineering, and economics.