integral of $\frac{1}{(1+e^{-x})}$

Question

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?

Answer 1

We have with $u=e^{-x}$ so $du=-e^{-x}dx\implies dx=-\frac{du}{u}$

$$\int \frac{dx}{1+e^{-x}}=-\int\frac{du}{u(1+u)}=\int\frac{du}{1+u}-\int\frac{du}{u}=\ln(1+u)-\ln u+C\\=\ln\left(1+e^x\right)+C$$

Answer 2

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 $$\dfrac{1}{s(s+1)} =-\left( \dfrac{1}{s}-\dfrac{1}{s-1}\right)$$

Answer 3

$$\int\frac{1}{1+e^{-x}}dx=\int\left[\frac{1}{1+e^{-x}}\times \frac{e^x}{e^x}\right]dx=\int\frac{e^x}{e^x+1}dx=\ln(e^x+1)+C$$

Answer 4

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: $$\int \frac {dx} {1+e^{-x}} =\int \frac 1 u \cdot - \frac {du}{u-1} = \int -\frac{du}{u(u-1)}$$

Answer 5

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 : $$ $$ $\int\frac{du}{u}=\ln(u)+C\\=\ln\left(1+e^x\right)+C$

Answer 6

$\int\frac{1}{1+e^{-x}}dx=\int\frac{e^x}{e^x+1}dx$

Substitute $e^x=u$

Then $e^xdx=du$

So

$\int\frac{du}{u+1}=ln(u+1)+C=ln(e^x+1)+C$