Evaluating the Definite Integral \( \int_1^3 (e^{\log x} + 1) dx \)

We are asked to evaluate the definite integral \( \int_1^3 (e^{\log x} + 1) dx \). This is a standard problem in calculus involving integration.

Simplifying the Integrand using Properties of Logarithms and Exponentials

First, let's simplify the expression inside the integral. We know a fundamental property relating exponentials and logarithms: for \(x > 0\), \(e^{\log x} = x\). Since our limits of integration are from 1 to 3, which are positive values, we can apply this property directly to the term \(e^{\log x}\).

So, the integrand \(e^{\log x} + 1\) simplifies to \(x + 1\).

The definite integral we need to evaluate becomes:

\(\displaystyle\int_1^3 (x + 1) dx\)

Finding the Antiderivative

Next, we find the antiderivative of the simplified integrand, which is \(x + 1\). The process of finding the antiderivative is part of basic integration techniques in mathematics.

Using the power rule for integration, \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\) (for \(n \ne -1\)) and the rule for integrating a constant, \(\int c dx = cx + C\), we find the antiderivative of each term:

• The antiderivative of \(x\) (where \(n=1\)) is \(\frac{x^{1+1}}{1+1} = \frac{x^2}{2}\).
• The antiderivative of \(1\) is \(1 \cdot x = x\).

So, the antiderivative of \((x + 1)\) is \(\frac{x^2}{2} + x\).

Applying the Limits of Integration using the Fundamental Theorem of Calculus

To evaluate the definite integral, we use the Fundamental Theorem of Calculus. This theorem states that if \(F(x)\) is an antiderivative of \(f(x)\), then the definite integral \(\int_a^b f(x) dx = F(b) - F(a)\). Here, our \(f(x) = x + 1\), our antiderivative \(F(x) = \frac{x^2}{2} + x\), the lower limit \(a=1\), and the upper limit \(b=3\).

We need to calculate \(F(3) - F(1)\).

First, evaluate \(F(3)\):

\(F(3) = \frac{3^2}{2} + 3 = \frac{9}{2} + 3\)

To add these, find a common denominator:

\(F(3) = \frac{9}{2} + \frac{3 \cdot 2}{2} = \frac{9}{2} + \frac{6}{2} = \frac{9+6}{2} = \frac{15}{2}\)

Next, evaluate \(F(1)\):

\(F(1) = \frac{1^2}{2} + 1 = \frac{1}{2} + 1\)

To add these, find a common denominator:

\(F(1) = \frac{1}{2} + \frac{1 \cdot 2}{2} = \frac{1}{2} + \frac{2}{2} = \frac{1+2}{2} = \frac{3}{2}\)

Now, subtract \(F(1)\) from \(F(3)\) to find the value of the definite integral:

\(\displaystyle\int_1^3 (x + 1) dx = F(3) - F(1) = \frac{15}{2} - \frac{3}{2}\)

\(= \frac{15 - 3}{2} = \frac{12}{2} = 6\)

Conclusion on the Definite Integral Evaluation

The value of the definite integral \( \int_1^3 (e^{\log x} + 1) dx \) is 6. This evaluation combines simplifying using properties of \(e^{\log x}\), finding the antiderivative through integration, and applying the limits using the Fundamental Theorem of Calculus, all key concepts in mathematics.

Thus, the result of this definite integral evaluation is 6.