Evaluate the integral using the given substitution. Integral of (dx)/(e^x + e^(-x)); u = e^x.

Question:

Evaluate the integral using the given substitution.

{eq}\int \frac{\mathrm{d}x}{e^x \, + \, e^{-x}} \, ; \; u = e^x {/eq}

Answer and Explanation:

Here the given integral is: {eq}\int \frac{\mathrm{d}x}{e^x \, + \, e^{-x}} {/eq}

Solving the integration by taking the following substitution:

{eq}u = {e^x} \Rightarrow du = {e^x}dx {/eq}

We get:

{eq}\begin{align*} I & = \int {\frac{{{\rm{d}}x}}{{{e^x}{\mkern 1mu} + {\mkern 1mu} {e^{ - x}}}}} \\ & = \int {\frac{{{\rm{d}}x}}{{\left( {{e^x}{\mkern 1mu} + {\mkern 1mu} \frac{1}{{{e^x}}}} \right)}}} \\ & = \int {\frac{{{e^x}{\rm{d}}x}}{{\left( {{e^{2x}} + 1} \right)}}} \\ & = \int {\frac{{du}}{{\left( {{u^2} + 1} \right)}}} \\ & = {\tan ^{ - 1}}\left( u \right) + c \\ & = {\tan ^{ - 1}}\left( {{e^x}} \right) + c \\ \end{align*} {/eq}


Explore our homework questions and answer library