Question

How do i solve this integral....

Answer 1

\[\int\limits_{-7}^{-3} \frac{ e^x }{ 1 - e^{2x} }dx\]

Answer 2

tried subbing u=e^x or e^2x and couldn't get anywhere

Answer 3

can you factor out an e^x?

Answer 4

i am going to butt in and guess that it is partial fractions

Answer 5

i was gonna say partials at some point :) but that just seemed to easy

Answer 6

\(\frac{1}{1-x^2}=\frac{1}{2}\frac{1}{1+x}-\frac{1}{2}\frac{1}{1-x}\) i think

Answer 7

but i could be wrong,

Answer 8

\[\frac{x}{1-x^2}\]

Answer 9

\[\frac{1}{1-x^2}\]

Answer 10

put \(x=e^x\)

Answer 11

hold the phone

Answer 12

\[\frac{ e^x }{ 1 - (e^{x})^2 }\to\~\frac{x}{1-x^2}\]

Answer 13

\[\Large \int\limits_{}^{}\frac{1}{1-t^2}dt=\int\limits_{}^{}\frac{-1}{1+t^2}dt\] is not right,
large, but not right

Answer 14

@amistre64 \(u=e^x, du=e^xdx\) gives
\[\int\frac{du}{1-u^2}\]

Answer 15

yup i did mistake :P

Answer 16

ahh i see, i was simply making a comparison for the partial decomp ..

Answer 17

hmm i wonder if it works that way too? probably does

Answer 18

thanks :)

Answer 19

partial fractions first, substitution second right?

Answer 20

yes

Answer 21

silly me

Answer 22

let t=e^x
dt=e^xdx

after
\[\Large \int\limits_{}^{}\frac{1}{1-t^2}dt\]
use the following.
\[\Large \int\limits_{}^{}\frac{1}{a^2-t^2}dt=\frac{1}{2a}\ln|\frac{a+t}{a-t}|+C\]