Attachment 26244this i tried to integrate by u substitution U=e^x+1 and then U-1/U after this i get U-lnU i dont know what do can anyone hint or help me solving it

Printable View

- Dec 15th 2012, 12:41 AMabdulrehmanshahlogarithm question need a slight help
Attachment 26244this i tried to integrate by u substitution U=e^x+1 and then U-1/U after this i get U-lnU i dont know what do can anyone hint or help me solving it

- Dec 15th 2012, 03:06 AMMarkFLRe: logarithm question need a slight help
Your substitution is good:

$\displaystyle u=e^x+1\,\therefore\,du=e^x\,dx$ giving us:

$\displaystyle \int_2^{e+1} \frac{1}{u}\,du$

Can you proceed from here?

Incidentally, the given answer is not quite correct.

By the way, I would post integration problems in the calculus forum. - Dec 15th 2012, 04:10 AMabdulrehmanshahRe: logarithm question need a slight help
not really u changed the question while i took it from a very reliable source

- Dec 15th 2012, 04:19 AMPlatoRe: logarithm question need a slight help
- Dec 15th 2012, 04:26 AMskeeterRe: logarithm question need a slight help
- Dec 15th 2012, 05:46 AMabdulrehmanshahRe: logarithm question need a slight help
ok the answer can be wrong but how can u change the limits provided by question can anyone help solving

- Dec 15th 2012, 05:56 AMskeeterRe: logarithm question need a slight help
when a substitution is done, one may "reset" the limits of integration from x-values to u-values ...

$\displaystyle u = 1 + e^x$

for $\displaystyle x = 0$ , $\displaystyle u = 1 + e^0 = 2$

for $\displaystyle x = 1$ , $\displaystyle u = e^1 + 1 = e + 1$

therefore ...

$\displaystyle \int_0^1 \frac{e^x}{1+e^x} \, dx = \int_2^{e+1} \frac{1}{u} \, du = \left[\ln{u}\right]_2^{e+1} = \ln(e+1) - \ln(2) = \ln\left(\frac{e+1}{2}\right)$ - Dec 15th 2012, 06:03 AMabdulrehmanshahRe: logarithm question need a slight help
this is the correct answer and as stated by my book so what is that is wrong about this question

- Dec 15th 2012, 06:07 AMskeeterRe: logarithm question need a slight help
**LOOK**at your "answer" again ...

http://mathhelpforum.com/attachments...lp-1111111.jpg - Dec 15th 2012, 06:08 AMabdulrehmanshahRe: logarithm question need a slight help
where did the e^x in the numerator go i know that lnu= 1/u but where did e^x go?

- Dec 15th 2012, 06:10 AMHallsofIvyRe: logarithm question need a slight help
I don't know what your book gave but

**you**posted the answer as $\displaystyle ln\left(\frac{1+ e^x}{2}\right)$. Since this is a definite integral there should be no "x" in the answer. The correct answer is, as skeeter said, $\displaystyle ln\left(\frac{1+ e}{2}\right)$ - Dec 15th 2012, 06:10 AMabdulrehmanshahRe: logarithm question need a slight help
ya i wrote it wrongly my bad the answer u got is the answer i have in the back of my note book can u please explain e^x part

- Dec 15th 2012, 06:13 AMskeeterRe: logarithm question need a slight help
$\displaystyle \int_0^1 \frac{{\color{red}e^x}}{1+e^x} \, {\color{red}dx} = \int_2^{1+e} \frac{{\color{red}du}}{u}$

$\displaystyle u = 1 + e^x$

$\displaystyle {\color{red}du = e^x \, dx}$