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Chapter 25: Problem 19

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

Short Answer

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

Step by step solution

01

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鈥檚 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, \( dx = \frac{du}{2e^{x}} \).
02

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: \[ \frac{1}{2} \int_{1}^{3} \frac{1}{u} du\]
03

Compute the Integral

This is now a straightforward integral. The antiderivative of \( \frac{1}{u} \) is \( \ln|u| \). Apply the fundamental theorem of calculus, \[ \frac{1}{2} [ \ln|u| ]_{1}^{3} = \frac{1}{2} (\ln 3 - \ln 1) \]
04

Substitute Back

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

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Most popular questions from this chapter

Compute the following integrals. $$ \int \frac{\sec ^{2}(\ln x)}{x} d x $$

It is April and, with the arrival of warm weather a pile of snow has turned into a puddle and the puddle is drying up. At time \(t=0\) there are 4 gallons of water in the puddle. The rate of change of water in the puddle is given by \(f(t)=-e^{0.2 t}\) gallons per hour. How much water is in the puddle \(t\) hours from now? When will the puddle dry up?

Find the given indefinite integral. $$ \int \frac{2 t}{\sqrt{2 t+6}} d t $$

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