91Ó°ÊÓ

The substitution technique is a powerful method in calculus integration that helps us to simplify complex integrals. The idea is to transform the original variable into a new one, making the integral easier to solve. In our example, the integral \( \int \frac{1}{1+e^x} \, dx \) is not in a straightforward form. Hence, substitution is a great approach.

We begin by letting \( u = 1 + e^x \). This substitution is strategic because the derivative of \( e^x \) is pairable with the differential \( du \). Thus, we express the original integral in terms of \( u \). Importantly:

• Derive \( du = e^x \, dx \).
• Solve for \( dx \) by isolating it: \( dx = \frac{du}{u-1} \).

This transforms the initial problem into an integrable form, providing a better path to solve the integral.

Partial fraction decomposition is a technique used to break down complex rational functions into simpler pieces that can be integrated individually. Once we've substituted and simplified the integrals, as seen in: \( \int \frac{1}{u(u-1)} \, du \), it's time for partial fraction decomposition.

This technique expresses \( \frac{1}{u(u-1)} \) as a sum of fractions. We write it as:

• \( \frac{1}{u(u-1)} = \frac{A}{u} + \frac{B}{u-1} \)
• This separates complex fractions into simpler linear ones.

To find constants \( A \) and \( B \), multiply both sides by the denominator, equate coefficients and solve:

• From \( 1 = A(u-1) + Bu \), solve \( A + B = 0 \) and \( -A = 1 \).
• The solution yields \( A = -1 \) and \( B = 1 \).

This essential step allows for easier term-by-term integration.

With both substitution and partial fraction decomposition techniques applied, it's time to evaluate the integral. Integrals can be definite or indefinite. Indefinite integrals result in a family of functions, expressed with a constant \( C \). In this step, each term is integrated separately:

For the terms resulting from partial fraction decomposition:

• \(-\int \frac{1}{u} \, du + \int \frac{1}{u-1} \, du \)

The solution involves basic natural logarithm integration rules:

• \(-\ln|u| + \ln|u-1| + C \)

Returning to the original variable yields the solution:

• \( \ln|\frac{e^x}{1+e^x}| + C \), simplified to \( x - \ln(1+e^x) + C \).

An indefinite integral indicates there's no specific interval for \( x \), focusing on the general solution. Definite integrals, while providing specific area values, require limits, which are not part of this specific example.