Need a little help starting this integral

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

I have tried substitution, partial fraction decomposition, and integration by parts but I can't seem to figure it out. Any help is much appriciated. o:)
 
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Hint: multiply top and bottom by e^{-x}.
 
okay, then I can use u sub, which nets...

-ln (1+e^{-x})

is that right?
 
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That does not seem right. What I have I done wrong?

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

u = 1+e^{-x}\\-du = e^{-x}

-\int\frac{1}{u}du
 
Last edited:
silverdiesel said:
That does not seem right. What I have I done wrong?

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

u = 1+e^{-x}\\-du = e^{-x}

-\int\frac{1}{u}du

It looks right to me, why do you think it's wrong?
 
After further review, I don't think its wrong. I guess I am just working on building confidence in the integration process and I was hoping for some confirmation that is was done right. Thank you. These forums have been extremely helpful and I really appreciate all the quick and intelligent responces.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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