Question

How to integrate lnabs(e^(x)+1)/e^x dx?

Answer 1

\[\frac{ \ln \left| e^x+1 \right| }{ e^x }dx\]

Answer 2

@hartnn @phi

Answer 3

how about \(u = e^{-x}\) ?
tried?

Answer 4

it works out :)
try it! and let me know if u get stuck

Answer 5

You mean the u-substitution?

Answer 6

yes

Answer 7

\(e^x+1 = \dfrac{1+e^{-x}}{e^{-x}}\)

Answer 8

How so?

Answer 9

\(e^x+1 = \dfrac{(e^x+1)e^{-x}}{e^{-x}}= \dfrac{1+e^{-x}}{e^{-x}}\)

Answer 10

But how does that help us to find the integral?
u=e^-x
du=-e^-x dx
And?

Answer 11

\(-\int \ln (\dfrac{u+1}{u})du\)

got this?

ln A/B = ln A-ln B

Answer 12

i am assuming we can use the standard integral of ln x

Answer 13

\(\int \ln x dx = x (\ln x -1)+c\)

Answer 14

just use that to get these :
\(\int \ln u dx -\int \ln (u+1) dx\)

Answer 15

if we cannot use it directly, we will need uv rule

Answer 16

Shouldn't we use the method of integration by parts in the first place?

Answer 17

trying that now...

\(\ln |e^x+1| (-e^{-x}) -\int \dfrac{e^x}{e^x+1}\times e^{-x}dx \)

yeah, works out...didn't find it shorter though

Answer 18

if we could directly use the result for ln x, then the u subs is easier/shorter.
otherwise, by parts from the beginning is easier.

Answer 19

Well, the result I got is \[-\ln \left| e^x+1 \right|e ^{-x}+x-\ln \left| e^x+1 \right|+C\]

Answer 20

\(\ln |e^x+1| (-e^{-x}) -\int \dfrac{e^{-x}}{e^{-x}+1}dx \\ u = 1+e^{-x}\)

Answer 21

\(\ln |e^x+1| (-e^{-x}) -\int \dfrac{e^{-x}}{e^{-x}+1}dx \\ u = 1+e^{-x} \\ \ln |e^x+1| (-e^{-x}) +\int \dfrac{1}{u}du \\\ln |e^x+1| (-e^{-x}) +\ln (e^{-x}+1)+c \)

Answer 22

did i make a sign mistake anywhere?

Answer 23

No.

Answer 24

http://www.wolframalpha.com/input/?i=integrate+ln%5Babs(e%5E(x)%2B1)%5D%2Fe%5Ex+dx

Answer 25

aah got it!

\(\ln |e^x+1| (-e^{-x}) -\int \dfrac{e^x}{e^x+1}\times (-e^{-x})dx\)

Answer 26

\(\ln |e^x+1| (-e^{-x}) +\int \dfrac{e^{-x}}{e^{-x}+1}dx \\ u = 1+e^{-x} \\ ...\)