Question

Preliminary steps The following integrals require a preliminary step such as a change of variables before using the method of partial fractions. Evaluate these integrals. $$\int \sqrt{e^{x}+1} d x \,(\text { Hint: Let } u=\sqrt{e^{x}+1}.)$$

Short answer

Answer: The result of the integral is $$\frac{1}{4}(e^x + 1) + C$$.

Step-by-step solution

Introduce substitution

Let \(u = \sqrt{e^x + 1}\). To find the relationship between \(du\) and \(dx\), we need to differentiate \(u\) with respect to \(x\).

Find the derivative

We first rewrite \(u^2 = e^x + 1\), and differentiate both sides with respect to \(x\): $$\frac{d}{dx}(u^2) = \frac{d}{dx}(e^x + 1)$$ Applying the Chain Rule, we get: $$2u \frac{du}{dx} = e^x$$ Now we'll solve for \(du\): $$\frac{du}{dx} = \frac{e^x}{2u}$$ So, \(du = \frac{e^x}{2u} dx\).

Replace in the integral

Now we substitute \(u\) and \(du\) into the original integral: $$\int \sqrt{e^x +1} dx = \int u \cdot \frac{e^x}{2u} dx = \frac{1}{2} \int u du$$

Integrate with respect to u

Now we can integrate the simplified expression: $$\frac{1}{2} \int u du = \frac{1}{2} \cdot \frac{1}{2} u^2 + C = \frac{1}{4}u^2 + C$$

Substitute back for x

Finally, we need to substitute the expression for \(u\) back in terms of \(x\): $$\frac{1}{4}u^2 + C = \frac{1}{4}(\sqrt{e^x + 1})^2 + C= \frac{1}{4}(e^x + 1) + C$$ So the final answer is: $$\int \sqrt{e^x +1} dx = \frac{1}{4}(e^x + 1) + C$$

Key concepts

Substitution Method

The substitution method is a powerful tool in calculus that simplifies the integration process by transforming an integral into a simpler form. This method involves replacing a complex expression with a new variable, typically denoted as \( u \).
This makes it easier to integrate. For example, in the given exercise, we have the integral \( \int \sqrt{e^x + 1} \, dx \).

• We start by letting \( u = \sqrt{e^x + 1} \). This step sets up the substitution.
• Next, we express \( dx \) in terms of \( du \). This involves differentiating \( u \) to find \( \frac{du}{dx} \).
• By modifying both sides, we can substitute back to achieve a simpler form of the integral.

By conducting this substitution, the integral \( \int \sqrt{e^x + 1} \, dx \) becomes \( \frac{1}{2} \int u \, du \).
This transformation is vital, as it makes the expression much easier to integrate, laying the groundwork for further steps in solving the integral.

Partial Fractions

While the integration by partial fractions is not directly applied in the provided task, understanding it as an integration technique is essential for tackling more complex problems.
Partial fraction decomposition turns a rational function into simpler fractions. These simpler fractions can then easily be integrated.

• First, ensure that the denominator's degree is greater than the numerator's. If not, perform polynomial division first.
• Next, decompose the rational function into partial fractions. This involves expressing it as a sum of fractions with simpler linear or quadratic denominators.
• Finally, integrate each fraction individually. This step often involves straightforward techniques or basic substitution.

Partial fractions provide a systematic way to handle integrals involving rational functions, offering a clear path to simplify and solve otherwise challenging integrals.
Though not used in our exercise directly, it's a technique worth knowing for integrals involving complex polynomial fractions.

Definite Integral Evaluation

Definite integral evaluation is the process of calculating the exact value of an integral over a specified interval. Although our exercise focuses on an indefinite integral, understanding how to extend this to definite integrals is useful.

• First, evaluate the antiderivative, or the indefinite integral, which is usually represented by a function \( F(x) \) plus a constant \( C \).
• Then, use the Fundamental Theorem of Calculus. It states that if \( F(x) \) is the antiderivative of \( f(x) \), then \( \int_a^b f(x) \, dx = F(b) - F(a) \).
• Replace \( F(x) \) with the calculated antiderivative and evaluate between bounds \( a \) and \( b \).

This evaluation provides the exact area under the curve \( f(x) \) from \( x = a \) to \( x = b \).
In practical terms, definite integral evaluation is crucial for finding real-world quantities, such as total distance, area, or mass, over specified ranges.