I'm stuck trying to solve this integral with "e". I am using the substitution method and made a sub, but cannot figure out how to apply it to this particular integral due to the sign of the exponent.
Let $\displaystyle \displaystyle u = e^{7x}$, then $\displaystyle \displaystyle \int \dfrac{dx}{4e^{-7x}+e^{7x}}\;{dx}$ $\displaystyle \displaystyle = \frac{1}{7}\int\frac{dx}{(\frac{4}{u}+u)u}\;{du} = \frac{1}{7}\int\frac{1}{4+u^2}\;{du} $. Let $\displaystyle \displaystyle t = 2\tan{u}$ to get $\displaystyle \displaystyle I = \frac{1}{14}\arctan\left(\frac{e^{7x}}{2}\right)+k$.