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Chapter 7: Problem 13

Evaluate the following integrals. $$\int \frac{e^{x}}{e^{x}+1} d x$$

Short Answer

Expert verified
Based on the given step-by-step solution, provide a short answer for evaluating the integral \(\int \frac{e^{x}}{e^{x}+1} d x\): The integral \(\int \frac{e^{x}}{e^{x}+1} d x\) can be evaluated using the substitution rule for integration with \(u=e^x + 1\). After performing the substitution and simplifying the integral, the result is \(\int \frac{e^{x}}{e^{x}+1} d x = \ln |e^x + 1| + C\), where C is the constant of integration.

Step by step solution

01

Substitute \(u\) for \(e^x+1\)

Let \(u=e^x + 1\). To find the differential of \(u\), write \(u\) in terms of \(x\) and then differentiate with respect to \(x\): $$ u = e^x + 1 \\ \frac{du}{dx} = \frac{d(e^x)}{dx}. $$ Since \(\frac{d(e^x)}{dx}= e^x\), we can rewrite this as: $$ d u = e^{x} d x. $$ Now, we can rewrite the integral in terms of \(u\) instead of \(x\): $$ \int \frac{e^x}{u} ( e^{x} d x)= \int \frac{1}{u} d u. $$
02

Integrate the \(u\)-integral

Now, we have simplified the integral to: $$ \int \frac{1}{u} d u. $$ This is a basic integral, since the integration of \(\frac{1}{u}\) is the natural logarithm of \(u\): $$ \int \frac{1}{u} d u = \ln |u| + C, $$ where \(C\) is the constant of integration.
03

Substitute back for \(x\)

Now we need to substitute the original expression of \(u\) in terms of \(x\), which was \(u=e^x + 1\). Therefore, the integral becomes: $$ \ln |e^x + 1| + C. $$ Hence, our final answer is: $$ \int \frac{e^{x}}{e^{x}+1} d x = \ln |e^x + 1| + C. $$

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