logarithm question need a slight help

**abdulrehmanshah** · December 15th 2012, 12:41 AM

this 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

**MarkFL** · December 15th 2012, 03:06 AM

Your substitution is good:

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

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

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.

**abdulrehmanshah** · December 15th 2012, 04:10 AM

not really u changed the question while i took it from a very reliable source

**Plato** · December 15th 2012, 04:19 AM

Originally Posted by abdulrehmanshah

not really u changed the question while i took it from a very reliable source

This is what you posted: $\int_0^1 {\frac{{e^x }}{{1 + e^x }}dx} = \mathop {\left. {\ln \left( {1 + e^x } \right)} \right|}\nolimits_{x = 0}^{x = 1}$

And the give answer is wrong.

**skeeter** · December 15th 2012, 04:26 AM

Originally Posted by abdulrehmanshah

not really u changed the question while i took it from a very reliable source

$\int_0^1 \frac{e^x}{e^x+1} \, dx = \left[\ln(e^x+1)\right]_0^1 = \ln(e+1) - \ln(2) = \ln\left(\frac{e+1}{2}\right)$

I agree with MarkFL2. I'd check that "source" again if I were you ...

**abdulrehmanshah** · December 15th 2012, 05:46 AM

ok the answer can be wrong but how can u change the limits provided by question can anyone help solving

**skeeter** · December 15th 2012, 05:56 AM

Originally Posted by abdulrehmanshah

ok the answer can be wrong but how can u change the limits provided by question can anyone help solving

when a substitution is done, one may "reset" the limits of integration from x-values to u-values ...

$u = 1 + e^x$

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

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

therefore ...

$\int_0^1 \frac{e^x}{1+e^x} \, dx = \int_2^{e+1} \frac{1}{u} \, du = \left[\ln{u}\right]_2^{e+1} = \ln(e+1) - \ln(2) = \ln\left(\frac{e+1}{2}\right)$

**abdulrehmanshah** · December 15th 2012, 06:03 AM

this is the correct answer and as stated by my book so what is that is wrong about this question

**skeeter** · December 15th 2012, 06:07 AM

Originally Posted by abdulrehmanshah

this is the correct answer and as stated by my book so what is that is wrong about this question

LOOK at your "answer" again ...

**abdulrehmanshah** · December 15th 2012, 06:08 AM

where did the e^x in the numerator go i know that lnu= 1/u but where did e^x go?

**abdulrehmanshah** · December 15th 2012, 06:10 AM

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

**HallsofIvy** · December 15th 2012, 06:10 AM

Originally Posted by abdulrehmanshah

this is the correct answer and as stated by my book so what is that is wrong about this question

I don't know what your book gave but you posted the answer as $ln\left(\frac{1+ e^x}{2}\right)$ . Since this is a definite integral there should be no "x" in the answer. The correct answer is, as skeeter said, $ln\left(\frac{1+ e}{2}\right)$

**skeeter** · December 15th 2012, 06:13 AM

$\int_0^1 \frac{{\color{red}e^x}}{1+e^x} \, {\color{red}dx} = \int_2^{1+e} \frac{{\color{red}du}}{u}$

$u = 1 + e^x$

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

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