I did this exercise but the book answer was =x-e^-x+c. And mine was

= -x-e^-x+c. Thanks for ur help.

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- Dec 2nd 2006, 06:24 AMron007Integration
I did this exercise but the book answer was =x-e^-x+c. And mine was

= -x-e^-x+c. Thanks for ur help. - Dec 2nd 2006, 07:00 AMCaptainBlack
- Dec 2nd 2006, 07:17 AMron007
- Dec 2nd 2006, 07:51 AMCaptainBlack
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 - Dec 2nd 2006, 08:01 AMron007
Thanks...