Nguyên hàm 8

Tìm $\displaystyle \int \dfrac{\ln(1+e^x)}{e^x}dx$

$\begin{aligned}D&=\dfrac{\ln(1+e^x)}{e^x}dx=-\ln(1+e^x)d\left(\dfrac{1}{e^x}\right)=-d\left(\dfrac{\ln(1+e^x)}{e^x}\right)+\dfrac{1}{e^x}d\left(\ln(1+e^x)\right)\\
&=-d\left(\dfrac{\ln(1+e^x)}{e^x}\right)+\dfrac{dx}{1+e^x}=-d\left(\dfrac{\ln(1+e^x)}{e^x}\right)+dx-\dfrac{e^xdx}{1+e^x}=-d\left(\dfrac{\ln(1+e^x)}{e^x}\right)+dx-d\left(\ln(1+e^x)\right)\end{aligned}$

Vậy $\displaystyle \int \dfrac{\ln(1+e^x)}{e^x}dx=-\dfrac{\ln(1+e^x)}{e^x}+x-\ln(1+e^x)+C$