Can someone tell me how to do an integral of this form?

[AxiomOfChoice]

The integral looks something like this:

<br /> \int \frac{dx}{e^x+1}.<br />

I'm interested in seeing the technique one should employ in analytically determining this indefinite integral, so please...no smart-alec answers like "put it in Mathematica." I tried that, and it wasn't particularly illuminating. Thanks.

 

[elect_eng]

The integral looks something like this:

<br /> \int \frac{dx}{e^x+1}.<br />

I'm interested in seeing the technique one should employ in analytically determining this indefinite integral, so please...no smart-alec answers like "put it in Mathematica." I tried that, and it wasn't particularly illuminating. Thanks.

Am i being a smart alec if I ask if you tried an integral table? I found it in mine. I'll think about how to solve it without a table.

 

[n!kofeyn]

The integral looks something like this:
<br /> \int \frac{dx}{e^x+1}.<br />

Use a u-substitution by letting u=e^x+1. Then solve for dx in terms of u and du. After this substitution, expand by partial fractions and use the natural logarithm to evaluate the integrals. If you are confused about how to execute these steps or need to see these steps actually written out, just let me know.

 

[arildno]

remember that \frac{du}{dx}=e^{x}=u-1\to{dx}=\frac{du}{u-1}

 

[n!kofeyn]

remember that \frac{du}{dx}=e^{x}=u-1\to{dx}=\frac{du}{u-1}

This is exactly what I had already posted.

 

[snipez90]

The integrand is also equivalent to 1 - [e^x / (e^x + 1)]. Then note that differentiating the denominator of [e^x / (e^x + 1)] gives e^x, which is the numerator.

 

[arildno]

This is exactly what I had already posted.

At least, that is definitely what your post implied!

I merely thought to give the OP an additional hint.

 

[adriank]

Another way to do it is to write

\int \frac{dx}{e^x+1} = \int \frac{e^{-x}}{1 + e^{-x}} \,dx.