I did this exercise but the book answer was =x-e^-x+c. And mine was
= -x-e^-x+c. Thanks for ur help.
There is virtualy nothing to show:
]$\displaystyle \int \frac{e^x+1}{e^x}dx$
divide through by the $\displaystyle e^x$ in the denominator to get:
$\displaystyle \int \frac{e^x+1}{e^x}dx=\int 1 + \frac{1}{e^x}dx=\int (1+e^{-x}) dx $
Distribute the integral over the addition:
$\displaystyle \int \frac{e^x+1}{e^x}dx=\int (1+e^{-x}) dx = \int 1\ dx + \int e^{-x} dx$
But the two integrals on the right are both standard integrals:
$\displaystyle \int 1\ dx=x+C$
$\displaystyle \int e^{-x} dx = -e^{-x}+K$
so:
$\displaystyle \int \frac{e^x+1}{e^x}dx=\int (1+e^{-x}) dx = \int 1\ dx + \int e^{-x} dx=x-e^{-x}+Const$
RonL