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JH

# Evaluate the integral.$\displaystyle \int \frac{1}{1 + 2e^x - e^{-x}}\ dx$

## $\frac{1}{3} \ln \left|2 e^{x}-1\right|-\frac{1}{3} \ln \left(e^{x}+1\right)+C$

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Integration Techniques

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### Video Transcript

Let's use a U substitution here. Let's take you to be either the ex then we know. Do you? Either the next T X and we could go out and solve this for X by dividing. So that's first. Let's rewrite. This is E D x. Oops as you DX since S is equal to you Some just replacing this with you here and then do you over you equals the X. So now, after using the use of you still have the one up there. Then that's one plus to you and then minus. And then here because of that minus sign well, one over you. And then here we multiply my DX do you over you. Now the next step years to just multiply this you into that denominator there. What? So that's what we have Could factor that denominator before you try partial fractions. And then here, you know the partial fractions will be of the form. Stay over to you minus one, and then be over you plus one. So, after finding in the way you find the end of the recall is you set this circle term equal to this circle term here, Then you multiply by both sides of this by the green circles from here and then saw for Andy. So after doing this, we end up with a is two thirds and then for being, we get minus one third. So I just pull off that minus. And then there's the you plus one and then here we know these in a girls if you need to, you could do it. Use up here. Here. You could do another U. So, in any case, one half and the natural log to you minus one. The one half is coming from the DW over, too. And then we also have minus one third and the natural log you plus one plus c. So just cancel those twos there. Let's go out and write this out. Cancel those twos. One over three. Also, I shouldn't circle this because we didn't back some in terms of X, new equals either the ex. So here we have one third Ellen to E X, minus one minus one third. And that here eating Ellen Ellen either the X plus one. No absolute values necessary here. This is a positive number. Plus he and that's your final answer

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Integration Techniques

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