# How do you integrate # 1/(1+e^x) # using partial fractions?

We use substitution to put the problem into a form where we can use partial fractions.

Rather than using partial fractions, we can also solve this with a little algebraic manipulation and substitution.

Substituting, we have

Plugging this into the equation above, we can get our result:

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To integrate ( \frac{1}{1+e^x} ) using partial fractions, we first express the fraction as a sum of simpler fractions. Since the denominator is a sum of two terms, (1) and (e^x), we write ( \frac{1}{1+e^x} ) as ( \frac{A}{1} + \frac{B}{e^x} ). Multiplying both sides by (1+e^x) to clear the denominators, we get ( 1 = A(e^x) + B(1) ).

To solve for ( A ) and ( B ), we let ( x = 0 ) to eliminate the term involving ( e^x ), yielding ( 1 = A + B ). Then, we let ( x = -\infty ) to eliminate the term involving ( 1 ), which gives us ( 1 = 0 + B ). Solving these equations simultaneously, we find that ( B = 1 ) and ( A = 0 ).

Thus, we have ( \frac{1}{1+e^x} = \frac{1}{e^x} ). Integrating both sides, we get ( \int \frac{1}{1+e^x} , dx = \int \frac{1}{e^x} , dx ).

The integral of ( \frac{1}{e^x} ) with respect to ( x ) is simply ( -e^{-x} + C ), where ( C ) is the constant of integration.

Therefore, ( \int \frac{1}{1+e^x} , dx = -e^{-x} + C ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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