# What is the integration of (log(1+e^x))/e^x ?

Sep 7, 2016

Let $I = \int \log \frac{1 + {e}^{x}}{e} ^ x \mathrm{dx}$

We will use the substn. ${e}^{x} = t \text{, so that, } {e}^{x} \mathrm{dx} = \mathrm{dt} , \mathmr{and} , \mathrm{dx} = \frac{\mathrm{dt}}{e} ^ x = \frac{\mathrm{dt}}{t}$.

$\therefore I = \int \log \frac{1 + t}{t} \cdot \frac{\mathrm{dt}}{t} = \int {t}^{-} 2 \log \left(1 + t\right) \mathrm{dt}$

Now, we use the Rule of Integration by Parts (IBP) :

$\text{(IBP) : } \int u v \mathrm{dt} = u \int v \mathrm{dt} - \int \left(\frac{\mathrm{du}}{\mathrm{dt}} \int v \mathrm{dt}\right) \mathrm{dt}$.

We take, $u = \log \left(1 + t\right) \Rightarrow \frac{\mathrm{du}}{\mathrm{dt}} = \frac{1}{1 + t}$, and,

$v = {t}^{-} 2 \Rightarrow \int v \mathrm{dt} = {t}^{- 2 + 1} / \left(- 2 + 1\right) = {t}^{-} \frac{1}{-} 1 = - \frac{1}{t}$.

$\therefore I = - \frac{1}{t} \log \left(1 + t\right) - \int \left\{\left(\frac{1}{1 + t}\right) \left(- \frac{1}{t}\right)\right\} \mathrm{dt}$

=-1/tlog(1+t)+int{1/((t(1+t))}dt

=-1/tlog(1+t)+int{(1+t-t)/((t(1+t))}dt

$= - \frac{1}{t} \log \left(1 + t\right) + \int \left\{\frac{1 + t}{t \left(1 + t\right)} - \frac{t}{t \left(1 + t\right)}\right\} \mathrm{dt}$

$= - \frac{1}{t} \log \left(1 + t\right) + \int \left\{\frac{1}{t} - \frac{1}{1 + t}\right\} \mathrm{dt}$

$= - \frac{1}{t} \log \left(1 + t\right) + \ln t - \ln \left(1 + t\right)$

=-1/e^xlog(1+e^x)+lne^x-ln(1+e^x)............".[as, "t=e^x]

$= - \log \frac{1 + {e}^{x}}{e} ^ x + x - \ln \left(1 + {e}^{x}\right)$

$= x - \left(\frac{1}{e} ^ x + 1\right) \ln \left(1 + {e}^{x}\right) + C$.

To be continued...