How Do I Evaluate $\int \frac{\mathrm{d}x}{e^x + 1} $?

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Evaluating integrals can sometimes feel like navigating a maze, especially when a straightforward substitution isn't immediately apparent. The integral ∫ dx/(e^x + 1) is a classic example where a bit of algebraic manipulation and a clever substitution can unlock the solution. This article will provide a detailed walkthrough of how to evaluate this integral, delving into the techniques and reasoning behind each step.

Initial Challenges and Strategic Approaches

When first encountering the integral ∫ dx/(e^x + 1), the presence of the exponential term in the denominator often suggests exploring u-substitution. However, a direct substitution of u = e^x doesn't immediately simplify the integral into a recognizable form. This is where strategic manipulation comes into play. We need to find a way to rewrite the integrand in a manner that facilitates a simpler substitution. One common approach is to multiply both the numerator and denominator by a suitable expression that allows us to isolate a term for substitution. In this case, multiplying by e^(-x) is a particularly effective strategy.

Multiplying the numerator and denominator by e^(-x) transforms the integral as follows:

∫ dx/(e^x + 1) = ∫ (e^(-x) dx) / (e(-x)(ex + 1)) = ∫ (e^(-x) dx) / (1 + e^(-x))

This transformation is crucial because it introduces the term e^(-x) in a way that makes a simple u-substitution viable. Now, let's explore the substitution process in detail.

The Power of U-Substitution: A Step-by-Step Solution

With the integral rewritten as ∫ (e^(-x) dx) / (1 + e^(-x)), the path to a solution becomes clearer. We can now employ the u-substitution method. Let's set:

u = 1 + e^(-x)

The next step is to find the differential du. Differentiating both sides with respect to x, we get:

du/dx = -e^(-x)

Therefore:

du = -e^(-x) dx

Notice that the numerator in our transformed integral contains e^(-x) dx, which is almost exactly what we have in our expression for du. We simply need to account for the negative sign. We can rewrite the integral in terms of u as follows:

∫ (e^(-x) dx) / (1 + e^(-x)) = - ∫ du / u

The integral -∫ du/u is a standard integral, and its solution is well-known:

  • ∫ du / u = - ln|u| + C

where C is the constant of integration. Now, we need to substitute back for u in terms of x:

  • ln|u| + C = - ln|1 + e^(-x)| + C

Since 1 + e^(-x) is always positive, we can drop the absolute value signs:

  • ln(1 + e^(-x)) + C

Thus, we have found one possible form of the integral. However, we can further simplify this expression to obtain a more elegant form.

Further Simplification and Alternative Forms

While - ln(1 + e^(-x)) + C is a correct solution, it's often desirable to express the result in a more simplified form. We can manipulate the logarithmic expression using properties of logarithms and algebraic techniques. Recall that e^(-x) = 1/e^x. Therefore:

  • ln(1 + e^(-x)) = - ln(1 + 1/e^x)

To combine the terms inside the logarithm, we can rewrite 1 as ex/ex:

  • ln(1 + 1/e^x) = - ln((e^x + 1)/e^x)

Now, we can use the logarithm property ln(a/b) = ln(a) - ln(b):

  • ln((e^x + 1)/e^x) = - [ln(e^x + 1) - ln(e^x)]

Since ln(e^x) = x, we have:

  • [ln(e^x + 1) - ln(e^x)] = - ln(e^x + 1) + x

Therefore, the integral can be expressed as:

∫ dx/(e^x + 1) = x - ln(e^x + 1) + C

This form is often preferred because it separates the transcendental and algebraic terms, providing a clearer representation of the solution. Let's summarize the steps we've taken and highlight the key concepts involved.

Summary of Steps and Key Concepts

To evaluate the integral ∫ dx/(e^x + 1), we followed these steps:

  1. Identified the Challenge: Recognizing that a direct u-substitution wasn't immediately obvious.
  2. Strategic Manipulation: Multiplying the numerator and denominator by e^(-x) to create a suitable form for substitution.
  3. U-Substitution: Setting u = 1 + e^(-x) and finding du = -e^(-x) dx.
  4. Integral Transformation: Rewriting the integral in terms of u as - ∫ du/u.
  5. Basic Integration: Evaluating - ∫ du/u as - ln|u| + C.
  6. Substitution Back: Replacing u with 1 + e^(-x) to get - ln(1 + e^(-x)) + C.
  7. Simplification: Using logarithm properties to simplify the result to x - ln(e^x + 1) + C.

The key concepts employed in this solution include:

  • U-Substitution: A fundamental integration technique that involves substituting a function and its derivative to simplify the integral.
  • Algebraic Manipulation: Rewriting the integrand to facilitate integration.
  • Logarithm Properties: Using properties of logarithms to simplify the result.

This example demonstrates the importance of strategic thinking and algebraic dexterity in integral calculus. Now, let's explore another method for solving this integral, showcasing the versatility of integration techniques.

Alternative Approach: Partial Fractions

Another way to approach the integral ∫ dx/(e^x + 1) is by using a method akin to partial fractions. While partial fractions are typically used for rational functions, we can adapt the idea to this integral by making a substitution that transforms the integrand into a rational function. Let's consider the substitution:

y = e^x

Then, dy/dx = e^x, so dx = dy/e^x = dy/y. Substituting these into the integral, we get:

∫ dx/(e^x + 1) = ∫ (dy/y) / (y + 1) = ∫ dy / (y(y + 1))

Now, we have a rational function that we can decompose into partial fractions. We want to find constants A and B such that:

1 / (y(y + 1)) = A/y + B/(y + 1)

Multiplying both sides by y(y + 1), we get:

1 = A(y + 1) + By

To solve for A and B, we can use strategic values of y. If y = 0, we get:

1 = A(0 + 1) + B(0) => A = 1

If y = -1, we get:

1 = A(-1 + 1) + B(-1) => B = -1

Thus, we can rewrite the integral as:

∫ dy / (y(y + 1)) = ∫ (1/y - 1/(y + 1)) dy

Now, we can integrate each term separately:

∫ (1/y - 1/(y + 1)) dy = ∫ (1/y) dy - ∫ (1/(y + 1)) dy = ln|y| - ln|y + 1| + C

Using logarithm properties, we can combine these terms:

ln|y| - ln|y + 1| + C = ln|y/(y + 1)| + C

Now, we substitute back y = e^x:

ln|y/(y + 1)| + C = ln|e^x / (e^x + 1)| + C

Since e^x and e^x + 1 are always positive, we can drop the absolute value signs:

ln(e^x / (e^x + 1)) + C

Using the logarithm property ln(a/b) = ln(a) - ln(b), we get:

ln(e^x / (e^x + 1)) + C = ln(e^x) - ln(e^x + 1) + C = x - ln(e^x + 1) + C

This result matches the simplified form we obtained using the previous method. This alternative approach demonstrates how different techniques can lead to the same solution, highlighting the interconnectedness of mathematical concepts.

Conclusion: Mastering Integration Techniques

Evaluating the integral ∫ dx/(e^x + 1) showcases the importance of strategic problem-solving in calculus. While a direct substitution might not always be apparent, techniques like algebraic manipulation, u-substitution, and methods akin to partial fractions can unlock the solution. We explored two distinct approaches: multiplying by a conjugate-like term and using u-substitution, and transforming the integral into a rational function suitable for partial fraction decomposition. Both methods led to the same simplified result:

∫ dx/(e^x + 1) = x - ln(e^x + 1) + C

This exercise reinforces the idea that mastering integration requires a versatile toolkit of techniques and a willingness to explore different avenues. By understanding the underlying principles and practicing various methods, you can confidently tackle a wide range of integral problems. Remember to always look for opportunities to simplify the integrand and consider different substitution strategies. The more you practice, the more intuitive these techniques will become, allowing you to approach integration challenges with greater confidence and skill.