integral of $\frac{1}{(1+e^{-x})}$ I make the substitution $u=1+e^{-x}$ which gives $-\dfrac{e^x}{u}\ du$.
Integrating gives me $$-e^x\ln(1+e^{-x}) + C,$$
but the answer is $\ln(e^x +1) + C$.
What am I doing wrong?
 A: We have with $u=e^{-x}$ so $du=-e^{-x}dx\implies dx=-\frac{du}{u}$
$$\int \frac{dx}{1+e^{-x}}=-\int\frac{du}{u(1+u)}=\int\frac{du}{1+u}-\int\frac{du}{u}=\ln(1+u)-\ln u+C\\=\ln\left(1+e^x\right)+C$$
A: Hint do another substitution. It is convenient to substitute only $e^u$ and not all the denominator.
$$\int\dfrac{1}{1+e^{-x}}\, dx = -\int\dfrac{1}{1+e^{u}}\, du$$
Now substitute $s=e^u, \dfrac{ds}{e^u}= du$ and use partial fraction decomposition
$$\dfrac{1}{s(s+1)} =-\left( \dfrac{1}{s}-\dfrac{1}{s-1}\right)$$
A: Your steps are not quite right. It seems you're skipping a couple of them.
From the substitution $u=1+e^{-x}$, you should get:
$$du=d(1+e^{-x}) = -e^{-x}dx$$
So that:
$$dx = - \frac {du}{e^{-x}}$$
Since $u=1+e^{-x}$ implies $e^{-x}=u-1$, this becomes:
$$dx = - \frac {du}{u-1}$$
So your integral should be:
$$\int \frac {dx} {1+e^{-x}} =\int \frac 1 u \cdot - \frac {du}{u-1} = \int -\frac{du}{u(u-1)}$$
A: $$\int\frac{1}{1+e^{-x}}dx=\int\left[\frac{1}{1+e^{-x}}\times \frac{e^x}{e^x}\right]dx=\int\frac{e^x}{e^x+1}dx=\ln(e^x+1)+C$$
A: Now, $\int\dfrac{1}{1+e^{-x}}dx$ = $\int\dfrac{e^{x}}{1+e^{x}}dx$. So letting $u= 1+e^{x}$, 
you get $$
$$ $du=e^{x} $ $dx$. $$
$$ so the integral becomes : $$
$$ $\int\frac{du}{u}=\ln(u)+C\\=\ln\left(1+e^x\right)+C$
A: $\int\frac{1}{1+e^{-x}}dx=\int\frac{e^x}{e^x+1}dx$
Substitute  $e^x=u$
Then $e^xdx=du$
So
$\int\frac{du}{u+1}=ln(u+1)+C=ln(e^x+1)+C$
