Solving the Antiderivative of 1/(e^x + 1)

[hypermonkey2]

I seem to be having trouble with this antiderivative can i please see a solution? Y= (1/(e^x + 1))

I tried substituting e^x + 1 as u, but that left me with the quest to find an antiderivative of 1/(u^2 - u). i feel like I am missing something silly here. thanks

 

[benorin]

You have it, keep going use partial fractions: \frac{1}{u(u-1)}= -\frac{1}{u}+\frac{1}{u-1}

 

[d_leet]

I seem to be having trouble with this antiderivative can i please see a solution? Y= (1/(e^x + 1))
I tried substituting e^x + 1 as u, but that left me with the quest to find an antiderivative of 1/(u^2 - u). i feel like I am missing something silly here. thanks

Try multiplying top and bottom by e^(-x)

 

[benorin]

Here it is (the integral)

For

\int\frac{dx}{e^x +1}

put u=e^x +1\Rightarrow x=\log (u-1) \mbox{ so that }dx=\frac{du}{u-1}

to get

\int\frac{dx}{e^x +1} = \int\frac{du}{u(u-1)} = \int\left( -\frac{1}{u}+\frac{1}{u-1}\right) du = -\log |u| + \log |u-1| + C
= -\log \left| e^x +1\right| + \log \left| e^x\right| + C = -\log \left| e^x +1\right| + x + C

 

[VietDao29]

For
\int\frac{dx}{e^x +1}
put u=e^x +1\Rightarrow x=\log (u-1) \mbox{ so that }dx=\frac{du}{u-1}
to get
\int\frac{dx}{e^x +1} = \int\frac{du}{u(u-1)} = \int\left( -\frac{1}{u}+\frac{1}{u-1}\right) du = -\log |u| + \log |u-1| + C
= -\log \left| e^x +1\right| + \log \left| e^x\right| + C = -\log \left| e^x +1\right| + x + C

?
Why on Earth are there still some people just post a complete solution (which is definitely, certainly, obviously, seriously, blah blah blah... against the forum's rules), without giving the OP an opportunity to solve the problem on his own?
Isn't your #2 post of this thread enough? Can't you wait for the OP to tell you if he can solve it or he still needs a little bit more help?
Am I really missing something?