Integrating $\int{\frac{1}{1+e^{x}}}dx$, Partial Fractions(?) [duplicate]

I need help with this integral:

$$H(x) = \int{\frac{1}{1+e^{x}}}dx$$

It should be easy, but I'm stuck. I thought about using a u-substitution but I didn't get any further. Am I meant to use partial fractions? I'm not yet very comfortable with partial fractions. I'd be thankful for someone's explanation!

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Another option to substitution, initially, is to multiply the numerator and denominator by $e^{-x}$, and then do a substitution. – Daryl May 15 '13 at 5:49

Let $e^x = t$. We then have $$\int \dfrac{dx}{1+e^x} = \int \dfrac{e^xdx}{e^x+e^{2x}} = \int \dfrac{dt}{t+t^2}$$

Equivalently, let $e^{-x} = t$, we then have $$\int \dfrac{dx}{1+e^x} = \int \dfrac{e^{-x}dx}{1+e^{-x}} = \int \dfrac{-dt}{1+t}$$

I trust you can take it forward using both/either of the above substitutions.

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I hate to disappoint, but I'm actually not sure what to do with $\int{\frac{dt}{t+t^{2}}}$. Could you give me a hint? – xisk May 15 '13 at 7:05
@xisk $$\dfrac{dt}{t+t^2} = \dfrac{dt}{t(1+t)} = \dfrac{dt}t - \dfrac{dt}{1+t}$$ Can you now finish it? – user17762 May 15 '13 at 7:06
Yes, I can! However, one question: why does $\frac{dt}{t(1+t)} = \frac{dt}{t} - \frac{dt}{1+t}$? Is that a use of partial fractions? – xisk May 15 '13 at 7:08
@xisk Yes, it is. If you put both over a common fraction, though, you will see $\frac1t-\frac1{1+t}=\frac{1+t}{t(1+t)}-\frac{t}{t(1+t)}=\frac1{t(1+t)}$ – Mike May 15 '13 at 8:38
@xisk Yes, it is the use of partial fractions. You can set $$\dfrac1{t(1+t)} = \dfrac{A}t + \dfrac{B}{1+t}$$ and find the coefficients $A$ and $B$. You will get them as $1$ and $-1$. – user17762 May 15 '13 at 16:05

Alternative way, $$\int \frac 1 {1+e^x}dx = \int \left(1-\frac{e^x}{1+e^x}\right)dx$$ Then use the substitution $t=1+e^x$.

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