I am trying to solve this integral and I need your suggestions.
I think about taking $1+e^{2x}$ and setting it as $t$, but I don't know how to continue now.

$$\int^1_0 \frac{dx}{1+e^{2x}}$$ Thanks!

  • 1
    $\begingroup$ Multiply the numerator and the denominator by $e^{-2x}$. $\endgroup$ – Random Variable May 14 '13 at 10:26

With the change of variable $u=e^{x}$, you get $$ \int_{[0,1]}\frac{dx}{1+e^{2x}} = \int_{[1,e]}\frac{1}{u(1+u^2)}du $$

  • $\begingroup$ Then expand in partial fraction, and it's ok. $\endgroup$ – Jean-Claude Arbaut May 14 '13 at 10:30
  • $\begingroup$ how its $u^2$ ? $\endgroup$ – Ofir Attia May 14 '13 at 11:03
  • $\begingroup$ What do you mean? ($e^{2x}=(e^{x})^2=u^2$, if that's what you ask) $\endgroup$ – Clement C. May 14 '13 at 12:10
  • $\begingroup$ @Clement C. after I get $\frac{1}{u(1+u^2)}du$ how I solve it with partial fraction? $\endgroup$ – Ofir Attia May 16 '13 at 7:46
  • $\begingroup$ By writing it as a sum of (easy to integrate) fractions, and integrating each of them independently. $\endgroup$ – Clement C. May 16 '13 at 11:53


Putting $e^{2x}=v\implies 2x=\ln v\implies 2dx=\frac{dv}v$

$$\int^1_0 \frac{dx}{1+e^{2x}}=\int_1^{ e^2}\frac1{2v(1+v)}dv=\frac12\cdot\int_1^{ e^2}\left(\frac1v-\frac1{v+1}\right)dv $$


$\int^{1}_{0}\frac{dx}{1+e^{2x}}=$ $ \int^{1}_{0}\frac{e^{-2x}dx}{1+e^{-2x}}=$ $-\frac{1}{2}\int^{1}_{0}\frac{(1+e^{-2x})'dx}{1+e^{-2x}}=$ $=-\frac{1}{2}ln(1+e^{-2x})|^{1}_{0}= \frac{1}{2}ln\frac{2e^{2}}{1+e^{2}}$


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