# Integration

• Dec 2nd 2006, 06:24 AM
ron007
Integration
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 AM
CaptainBlack
Quote:

Originally Posted by ron007
I did this exercise but the book answer was =x-e^-x+c. And mine was
= -x-e^-x+c. Thanks for ur help.

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

As a check you can try differentiation this and your solution

RonL
• Dec 2nd 2006, 07:17 AM
ron007
Quote:

Originally Posted by CaptainBlack
$\int \frac{e^x+1}{e^x}dx=\int (1+e^{-x}) dx =x - e^{-x}+Const$

As a check you can try differentiation this and your solution

RonL

Thanks. But can u show the process to see my error.
• Dec 2nd 2006, 07:51 AM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
$\int \frac{e^x+1}{e^x}dx=\int (1+e^{-x}) dx =x - e^{-x}+Const$

As a check you can try differentiation this and your solution

RonL

Quote:

Originally Posted by ron007
Thanks. But can u show the process to see my error.

There is virtualy nothing to show:

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

divide through by the $e^x$ in the denominator to get:

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

Distribute the integral over the addition:

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

But the two integrals on the right are both standard integrals:

$\int 1\ dx=x+C$

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

so:

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

RonL
• Dec 2nd 2006, 08:01 AM
ron007
Thanks...