MHB Solving a Tricky Integral: 1/[3+e^(-x)]

[filiphenrique]

Hi everyone, I had a test last week, and I couldn't solve this integral: 1/[3+e^(-x)]
How do I solve this one? Thanks in advance.

 

[Ackbach]

Try $u=3+e^{-x}$, so $u-3=e^{-x}$, and $du=-e^{-x} \, dx$. The integral becomes
$$-\int\frac{1}{u} \, \frac{du}{u-3}.$$
Can you take it from here?

 

[filiphenrique]

Hi. I don't know if it's me that don't get what you've done, or if the the integral I wrote is clear. Here it is: View attachment 2392. The way you start solving is the same to this integral?

 

[Ackbach]

Yes, I assumed you started with
$$\int \frac{1}{3+e^{-x}} dx.$$
The $u$-substitution I outlined above will give you the new integral
$$-\int\frac{du}{u(u-3)}.$$
Can you proceed from here?

 

[Prove It]

Hi everyone, I had a test last week, and I couldn't solve this integral: 1/[3+e^(-x)]
How do I solve this one? Thanks in advance.

I'd probably do this:

$\displaystyle \begin{align*} \int{ \frac{1}{3 + e^{-x}} \, dx } &= \int{ \frac{e^x}{e^x \left( 3 + e^{-x} \right) } \, dx } \\ &= \int{ \frac{e^x}{3e^x + 1} \, dx } \\ &= \frac{1}{3} \int{ \frac{3e^x}{3e^x + 1} \, dx } \end{align*}$

Now make the substitution $\displaystyle \begin{align*} u = 3e^x + 1 \implies du = 3e^x\,dx \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \frac{1}{3} \int{ \frac{3e^x}{3e^x + 1} \, dx} &= \frac{1}{3} \int{ \frac{1}{u}\,du } \\ &= \frac{1}{3}\ln{ |u| } + C \\ &= \frac{1}{3} \ln{ \left| 3e^x + 1 \right| } + C \\ &= \frac{1}{3} \ln{ \left( 3e^x + 1 \right) } + C \textrm{ since } 3e^x + 1 > 0 \textrm{ for all } x \end{align*}$