Integrate (e^-x - e^x)/(e^2x + e^-2x +2) dx

Question

zzr0ck3r · Answer

start with u = e^x
then you will have partial fractions to deal with, then more substitutions on each term

anonymous · Answer

let u=e^x+e^(-x)

anonymous · Answer

$$u=e^x+e^{-x} \ \ \ then \ \ \ du=(e^x-e^{-x})dx \ e^{2x}+e^{-2x}+2=(e^x+e^{-x})^2=u^2 \ so \ \ I=\int\limits \frac{-du}{u^2}$$

anonymous · Answer

notice  e^-2x+e^-2x+2=(e^x+e^-x)^2  
also
(e^-x-e^x)/(e^x+e^-x)^2=(1/2)((e^-x-e^x)/2/((e^x+e^-x)^2)/4)
=(1/2)(-(e^x-e^-x)/2/(coshx)^2)=(1/2)(-sinh x/(cosh x)^2)
so we have:
$$-1/2\int\limits sinhx/(coshx)^2dx$$
use sub. u=coshx  du=sinhxdx so
$$-1/2\int\limits 1/u^2du=1/2u+c=1/2\cosh(x) + c$$

anonymous · Answer

$$I=\int\limits \frac{-du}{u^2}=\frac{1}{u}+c=\frac{1}{e^x+e^{-x}}+c=\frac{1}{2 \cosh x}+c$$

anonymous · Answer

I really feel like if he needs help with this hyperbolic trig functions aren't going to help...most people have trouble enough with ordinary trig functions lol...