How Do You Solve This Integral with \( e \) in the Denominator?

[tmt1]

How can I go about solving this problem?

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

I am unsure how to start.

 

[MarkFL]

I would begin by multiplying the integrand by:

\(\displaystyle 1=\frac{e^x}{e^x}\)

and then use the substitution:

\(\displaystyle u=e^x\)

Can you proceed?

 

[tmt1]

I would begin by multiplying the integrand by:

\(\displaystyle 1=\frac{e^x}{e^x}\)

and then use the substitution:

\(\displaystyle u=e^x\)

Can you proceed?

so I would get

$$\int_{}^{} \frac{e^x}{2} \,dx$$

Then $e^x = u$ so $du = e^x dx$. Then I can substitute that and get

$$\frac{1}{2} \int_{}^{} \,du$$

Or $\frac{1}{2} u + C$ which becomes $\frac{1}{2} e^x + C$. Is this right?

 

[MarkFL]

so I would get

$$\int_{}^{} \frac{e^x}{2} \,dx$$

Then $e^x = u$ so $du = e^x dx$. Then I can substitute that and get

$$\frac{1}{2} \int_{}^{} \,du$$

Or $\frac{1}{2} u + C$ which becomes $\frac{1}{2} e^x + C$. Is this right?

No, let's begin with the original:

\(\displaystyle I=\int\frac{1}{e^x+e^{-x}}\,dx\)

Now, using my first suggestion, we obtain:

\(\displaystyle I=\int\frac{1}{e^x+e^{-x}}\cdot\frac{e^x}{e^x}\,dx=\int\frac{e^x}{\left(e^x\right)^2+1}\,dx\)

Then, using my second suggestion:

\(\displaystyle u=e^x\implies du=e^x\,dx\)

we now have:

\(\displaystyle I=\int\frac{1}{u^2+1}\,du\)

Now you can proceed either by recognizing this is a well-known integral, or use a trigonometric substitution. :)

 

[soroban]

$$\int \frac{dx}{e^x + e^{-x}}$$

Careful!

MarkFL's advice goes like this:

$$\int \frac{e^x}{e^x}\cdot\frac{1}{e^x + e^{-x}}\,dx \;=\;\int \frac{e^x}{e^{2x} + 1}\,dx \;=\;\int \frac{e^x}{(e^x)^2 + 1}\,dx $$

$$\text{Let }u = e^x \quad\Rightarrow\quad du = e^x\,dx $$

$$\text{Substituite: }\;\int \frac{du}{u^2+1} $$

$\text{Got it?}$