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I am having troubles with this integral:

$$\int \frac{1}{\sqrt{e^{2x}+e^x+1}}dx.$$

Could anyone help me?

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closed as off-topic by Carl Mummert, heropup, Krish, Aaron Maroja, Najib Idrissi Mar 30 '15 at 20:22

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    $\begingroup$ Try to substitute $u = e^x$. $\endgroup$ – user99914 Mar 30 '15 at 9:04
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$\bf{My\; Solution}::$ We can write it as $\displaystyle \int\frac{e^{-x}}{\sqrt{1+e^{-x}+e^{-2x}}}dx\;$

Now Let $e^{-x}=t\;,$ Then $e^{-x}dx = -dt$. So our integral becomes $\displaystyle - \int\frac{1}{\sqrt{t^2+t+1}}dt$

Which is equal to $\displaystyle = -\int\frac{1}{\sqrt{\left(t+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}}dt$

Now Using $\displaystyle \int\frac{1}{\sqrt{x^2+a^2}}dx = \ln \left|x+\sqrt{x^2+a^2}\right|+\mathcal{C}$

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  • $\begingroup$ Your very first integral's central summand in the denominator must be $\;e^{-x}\;$ $\endgroup$ – Timbuc Mar 30 '15 at 10:34
  • $\begingroup$ Are you sure about the substitution? Shouldnt the integral after substitution be: - int t / sqrt( t^2 + t + 1 )*( dt / e ^ -x ) ? $\endgroup$ – sykatch Mar 30 '15 at 18:52
  • $\begingroup$ Or is it possible not to substitute every e ^ -x with t so you get: - int e ^ -x / sqrt( t^2 + t + 1 )*( dt / e ^ -x ) $\endgroup$ – sykatch Mar 30 '15 at 19:41

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