An Integral with a Fraction and e

[tmt1]

$$ \int_{}^{} \frac{1}{ e^x + {e}^{-x}}\,dx $$

I have this integral, and I'm not sure how to approach it. I tried u-substitution with $u = e^x + {e} ^{-x}$, but that seemed to go to a dead end. I'm not sure how to apply partial fractions to this problem. Is there a better way?

 

[Jameson]

Hi tmt,

I would try multiplying by $$\frac{e^x}{e^x}$$ and then using the substitution $u=e^x$. That should get it into a form where you can use trig substitution to solve. :)

 

[Prove It]

Alternatively, since $\displaystyle \begin{align*} \cosh{(x)} = \frac{1}{2}\,\left( \mathrm{e}^x + \mathrm{e}^{-x} \right) \end{align*}$ that means the integral is

$\displaystyle \begin{align*} \int{\frac{1}{\mathrm{e}^x + \mathrm{e}^{-x}}\,\mathrm{d}x} &= \int{ \frac{1}{2\cosh{(x)}}\,\mathrm{d}x } \\ &= \int{ \frac{\cosh{(x)}}{2\cosh^2{(x)}}\,\mathrm{d}x} \\ &= \frac{1}{2} \int{ \frac{\cosh{(x)}}{1 + \sinh^2{(x)}}\,\mathrm{d}x} \end{align*}$

You can now substitute $\displaystyle \begin{align*} u = \sinh{(x)} \implies \mathrm{d}u = \cosh{(x)}\,\mathrm{d}x \end{align*}$.