# Thread: Stuck integrating with "e"

1. ## Stuck integrating with "e"

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.

2. Can't you say the following,

$\int (\frac{1}{4}e^{7x} + \frac{1}{e^{7x}})dx$

EDIT: Nevermind, this is incorrect, I'll edit this post if I find something else.

Harish beat me to it!

3. $4e^{-7x}+e^{7x} = \dfrac{4}{e^{7x}}+e^{7x}= \dfrac{4+e^{2(7x)}}{e^{7x}}$

$\int \dfrac{dx}{4e^{-7x}+e^{7x}} = \int \dfrac{e^{7x}}{4+e^{2(7x)}} dx$

now let $u = e^{7x}$ and complete

4. Thank you this helped me solve it and was able to verify with Pandevil1990's scan.

5. Let $\displaystyle u = e^{7x}$, then $\displaystyle \int \dfrac{dx}{4e^{-7x}+e^{7x}}\;{dx}$ $\displaystyle = \frac{1}{7}\int\frac{dx}{(\frac{4}{u}+u)u}\;{du} = \frac{1}{7}\int\frac{1}{4+u^2}\;{du}$. Let $\displaystyle t = 2\tan{u}$ to get $\displaystyle I = \frac{1}{14}\arctan\left(\frac{e^{7x}}{2}\right)+k$.