Problem 19 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Compute the following integrals. $$ \int_{0}^{\ln 2} \frac{e^{x}}{e^{2 x}+1} d x $$

Step-by-Step Solution

Answer

\(\frac{1}{2} (\ln 3 - \ln 2)\)

Step 1: Set Up the Substitution

Let, \( u = e^{2x}+ 1 \). Compute the derivative of \( u \) so we can replace \( dx \) in the integral. The derivative \( du = 2e^{2x} dx \). Here, we notice that we don't have \( 2e^{2x} \) in the given problem, but there’s a similar term, \( e^{x} \). We could revise our \( u \) to suit better to the given problem, but by dividing by \( 2e^{x} \) we make it fit. Hence, \( dx = \frac{du}{2e^{x}} \).

Step 2: Perform the Substitution

Substitute \( u \) and \( dx \) into our original integral: \[ \int_{1}^{3} \frac{e^{x}}{u} \cdot \frac{du}{2e^{x}} \]After cancellation of \( e^{x} \) in both numerator and denominator, this simplifies to: \[ \frac{1}{2} \int_{1}^{3} \frac{1}{u} du\]

Step 3: Compute the Integral

This is now a straightforward integral. The antiderivative of \( \frac{1}{u} \) is \( \ln|u| \). Apply the fundamental theorem of calculus, \[ \frac{1}{2} [ \ln|u| ]_{1}^{3} = \frac{1}{2} (\ln 3 - \ln 1) \]

Step 4: Substitute Back

Substitute back our original expression for \( u = e^{2x} + 1 \). \[ \frac{1}{2} (\ln (2 e^{\ln 2} + 1) - \ln (e^0 + 1)) = \frac{1}{2} (\ln 3 - \ln 2)) \]