Question

\int\limits_0^1 (dx/e^x+e^-x)

Answer 1

start by multiplying top and bottom wid \(\large e^x\)

Answer 2

\(\mathbb{\int_0^1 \frac{1}{e^x+e^{-x}}dx}\)

\(\large \mathbb{\int_0^1 \frac{e^x}{e^x[e^x+e^{-x}]}dx}\)

Answer 3

Can u show me the process? i am not getting anything :) yeah then what?

Answer 4

\(\large \mathbb{\int_0^1 \frac{1}{e^x+e^{-x}}dx}\)

\(\large \mathbb{\int_0^1 \frac{e^x}{e^x[e^x+e^{-x}]}dx}\)

\(\large \mathbb{\int_0^1 \frac{e^x}{(e^x)^2+e^{-x+x}}dx}\)

\(\large \mathbb{\int_0^1 \frac{e^x}{(e^x)^2+e^{0}}dx}\)

\(\large \mathbb{\int_0^1 \frac{e^x}{(e^x)^2+1}dx}\)

Answer 5

see if that looks okay,
next u need to substitute and pull an inverse integral formula..

Answer 6

so i have to let z=e^x and then d/dx e^x=dz/dx
=>e^x dx=dz
and then by doing substitution i get
i get (tan^-1z)

Answer 7

thank you ganeshie :)

Answer 8

just one more question is e^1=e and e^0=1? :)

Answer 9

yup ! you got it !!!!!

Answer 10

thank you for the help :)

Answer 11

genius :)

Answer 12

The ultimate genius mam :)

Answer 13

\(\large \mathbb{\int_0^1 \frac{1}{e^x+e^{-x}}dx}\)

\(\large \mathbb{\int_0^1 \frac{e^x}{e^x[e^x+e^{-x}]}dx}\)

\(\large \mathbb{\int_0^1 \frac{e^x}{(e^x)^2+e^{-x+x}]}dx}\)

\(\large \mathbb{\int_0^1 \frac{e^x}{(e^x)^2+e^{0}]}dx}\)

\(\large \mathbb{\int_0^1 \frac{e^x}{(e^x)^2+1]}dx}\)


sub \(\large z = e^x \)

\(\large dz = e^x dx \)


\(\large \mathbb{\int_0^1 \frac{1}{(z)^2+1]}dz}\)

Answer 14

lol you're the genius :)

Answer 15

I don't know anything about anything mam :) thank you for the help again mam :)

Answer 16

\(\large \mathbb{\int_0^1 \frac{1}{e^x+e^{-x}}dx}\)

\(\large \mathbb{\int_0^1 \frac{e^x}{e^x[e^x+e^{-x}]}dx}\)

\(\large \mathbb{\int_0^1 \frac{e^x}{(e^x)^2+e^{-x+x}]}dx}\)

\(\large \mathbb{\int_0^1 \frac{e^x}{(e^x)^2+e^{0}]}dx}\)

\(\large \mathbb{\int_0^1 \frac{e^x}{(e^x)^2+1]}dx}\)


sub \(\large z = e^x \)

\(\large dz = e^x dx \)


\(\large \mathbb{\int_0^1 \frac{1}{(z)^2+1]}dz}\)

\(\large \mathbb{ \tan^{-1}z\Big|_0^1}\)

\(\large \mathbb{ \tan^{-1}(e^x)\Big|_0^1}\)

Answer 17

\(\large \mathbb{\int_0^1 \frac{1}{e^x+e^{-x}}dx}\)

\(\large \mathbb{\int_0^1 \frac{e^x}{e^x[e^x+e^{-x}]}dx}\)

\(\large \mathbb{\int_0^1 \frac{e^x}{(e^x)^2+e^{-x+x}]}dx}\)

\(\large \mathbb{\int_0^1 \frac{e^x}{(e^x)^2+e^{0}]}dx}\)

\(\large \mathbb{\int_0^1 \frac{e^x}{(e^x)^2+1]}dx}\)


sub \(\large z = e^x \)

\(\large dz = e^x dx \)


\(\large \mathbb{\int_0^1 \frac{1}{(z)^2+1]}dz}\)

\(\large \mathbb{ \tan^{-1}z\Big|_0^1}\)

\(\large \mathbb{ \tan^{-1}(e^x)\Big|_0^1}\)

\(\large \mathbb{ \tan^{-1}(e^1) -\tan^{-1}(e^0) }\)

Answer 18

that simplifies to :

\(\large \mathbb{ \tan^{-1}(e) -\tan^{-1}(1) }\)

\(\large \mathbb{ \tan^{-1}(e) -\frac{\pi}{4}}\)

Answer 19

i can see u covered so much for one day :) u should relax sometime :)

Answer 20

no i didn't finish anything yet..have exam on wednesday..didn't study.just doing math homework :) thank you again ...i am lazy :)

Answer 21

ohhwelcome to the club of lazy ppl xD,
there is an well established theary -- all smart+efficient ppl are lazy :P

Answer 22

may i knw u in wat grade ha ?

Answer 23

But i am dull also lol :) i am not smart enough like you all Openstudy genius students :)

Answer 24

I am in university lol..that's why i am saying i am dull lol :)

Answer 25

then its easy for u to prepare :)
3 days is sufficient for u to prepare for exams im sure... good luck !!

Answer 26

thank you mam :)

Answer 27

np.. .tag me in ur future questions :)

Answer 28

I will thanks :)

Answer 29

i wish i could give thousand medels to u :) such a good person and nice person like u :)

Answer 30

you're sweet :3

Answer 31

:) u are more sweet i am cruel lol :) thanks