Question

How do you solve this integral? I tried many times but it doesn't work... I want to use the integration by parts.

Answer 1

when substituting, i think you may have forgotten to substitute e^x

Answer 2

Were you referring to the notes that were written on the question? because I know it's correct and I double-checked multiple times but the integration seems to go on forever..

Answer 3

http://www.wolframalpha.com/input/?i=integral+from+ln%283%29+to+ln%288%29+of+%28xe%5Ex%29%2F%28sqrt+of+%281%2Be%5Ex%29%29

Answer 4

is this the answer ?

Answer 5

int by parts might be it

derive x and integrate the rest

Answer 6

\[\int x\ \frac{e^x}{(1+e^x)^{1/2}}dx=x\ 2(1+e^x)^{1/2}-\int 2(1+e^x)^{1/2}dx \]

Answer 7

looks weird in wolfram

Answer 8

the answer is -4+20(In2)-6In3

Answer 9

@amistre64 Yes, I used the integration by parts but it went on forever..

Answer 10

I did u = 1 + e^x though

Answer 11

then what does x equal in terms of u?

Answer 12

x = Inu, I think?

Answer 13

if we let u=e^x; then dx = du/u and x = ln(u) would be right

Answer 14

if u = 1+e^2; then x=ln(u-1)

Answer 15

Yes, I'm following

Answer 16

u = 1+e^x
du = e^x dx ; dx = du/e^x

\[\frac{xe^x}{u^{1/2}e^x}du\]

\[\frac{x}{u^{1/2}}du\]
x = ln(u-1)

\[\frac{ln(u-1)}{u^{1/2}}du\]

might be workable

Answer 17

\[\int ln(u-1)u^{-1/2}du=ln(u-1)\ 2u^{1/2}-2\int \frac{u^{1/2}}{(u-1)}du\]

Answer 18

oh yeah, the wolf proves this thing is not just a simple working

Answer 19

So do I do integration by parts again for the second term?

Answer 20

Cause I did that and it went on forever

Answer 21

the wolf starts out using u = sqrt(1+e^x)

Answer 22

du = e^x/2sqrt(1+e^x) dx

2sqrt(1+e^x)/e^x du = dx

2u/e^x du = dx

\[\frac{2uxe^x}{ue^x}du=2xdu\]

Answer 23

u = sqrt(1+e^x)

u^2 = 1+e^x

ln(u^2-1) = x

Answer 24

u^2-1 = (u-1)(u+1)

ln((u-1)(u+1)) = ln(u-1) + ln(u+1)

\[\int 2ln(u-1)+\int 2ln(u+1)\]

maybe

Answer 25

Thanks, I'll try that