Question

find the integral of (1+e^x)/(1-e^x)

Answer 1

(1+e^x)(1-e^x)=1-e^2x
now integrating we get
x-e^2x/2+c

Answer 2

can you pls show me the step by step procedure?

Answer 3

i think you forgot the fraction bar between the two quantities..

Answer 4

oh sorry

Answer 5

please help me

Answer 6

(1+e^x)/(1-e^x)
\[1/(1-e^x)+e^x/(1-e^x)\]
now integrate both terms one by one
in first term we divide numerator and denominator by e^x we get
\[e^-x/e^-x-1\]
now if we take (e^-x-1) as t then -e^-xdx=dt
so e^-xdx=-dt
we are left for first term as -dt/t
integrate it we -ln(t)
t=e^-x-1
so integration of first we get -ln(e^-x-1)

Answer 7

now second term is e^x/(1-e^x)
if we take 1-e^x=t then e^xdx=-dt
so dt/t on integrating gives -ln (t)
so second term's integration gives -ln(1-e^x)
so complete answer is
-ln(e^-x-1)-ln(1-e^x)+c

Answer 8

did you get it?

Answer 9

yes, thanks..I'm a little confused about that problem a while ago, thanks for the help

Answer 10

\[ \int \frac{1+e^x}{1-e^x}\]
\[\int \frac{1}{1-e^x}+\frac{e^x}{1-e^x}\]
\[\int \frac{e^{-x}}{e^{-x}-1}+\frac{e^x}{1-e^x}\]
\[-ln(e^{-x}-1)-ln(1-e^{x})+C\]