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To integrate 1/(1+e^x), use substitution u = 1+e^x and simplify using partial fractions.

To integrate 1/(1+e^x), start by using substitution u = 1+e^x. This gives du/dx = e^x, so dx = du/e^x. Substituting these into the integral gives:

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

Next, use partial fractions to simplify the integrand. Write 1/u as A/(1+e^x) + B/(1+e^(-x)). Then:

1/u = A/(1+e^x) + B/(1+e^(-x))

1 = A(1+e^(-x)) + B(1+e^x)

Let x = 0: 1 = A(2) + B(2) => A + B = 1/2

Let x = ln(2): 1 = A(1/2) + B(2) => A = -1/2, B = 3/2

Substituting these values back into the integral gives:

∫ (1/u) * e^(-x) du = ∫ (-1/2)/(1+e^x) + (3/2)/(1+e^(-x)) dx

= (-1/2)ln(1+e^x) + (3/2)ln(1+e^(-x)) + C

Therefore, the integral of 1/(1+e^x) is (-1/2)ln(1+e^x) + (3/2)ln(1+e^(-x)) + C.

For more detail on the strategies used in integration and to explore different techniques, you can visit our detailed page on `Techniques of Integration`

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** A-Level Maths Tutor Summary:** To integrate 1/(1+e^x), we start with a substitution to simplify the expression, and then apply partial fractions to break it down further. By setting u = 1+e^x, we find that the integral equals (-1/2)ln(1+e^x) + (3/2)ln(1+e^-x) + C. This method combines substitution and partial fractions for solving the integral.

To understand more about the context and use of indefinite and definite integrals in calculus, refer to our comprehensive notes on `Definite and Indefinite Integrals`

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Additionally, for a complete list of integration rules that can be helpful in solving similar problems, please check our section on `Integration Rules`

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