L lethalasian New member Joined Feb 14, 2020 Messages 15 Mar 31, 2020 #1 integral e^x / sqrt(1-e^(2x)) DX The chapter I am on is strategies for integration. Would I substitute e^x = U and then go on to do (1-u^2) using trig substitution? I am stuck on this question any help would be appreciated

integral e^x / sqrt(1-e^(2x)) DX The chapter I am on is strategies for integration. Would I substitute e^x = U and then go on to do (1-u^2) using trig substitution? I am stuck on this question any help would be appreciated

pka Elite Member Joined Jan 29, 2005 Messages 9,809 Mar 31, 2020 #2 This is rather straightforward. We get \(\displaystyle\int {\frac{1}{{\sqrt {1 - {u^2}} }}du = \arctan (u)} \)

This is rather straightforward. We get \(\displaystyle\int {\frac{1}{{\sqrt {1 - {u^2}} }}du = \arctan (u)} \)

Jomo Elite Member Joined Dec 30, 2014 Messages 7,038 Mar 31, 2020 #3 You need to be complete. If u =e^x, then du =e^xdx. This yields the integral posted by pka which is supposed to be known to you. Last edited: Mar 31, 2020

You need to be complete. If u =e^x, then du =e^xdx. This yields the integral posted by pka which is supposed to be known to you.

topsquark Full Member Joined Aug 27, 2012 Messages 920 Mar 31, 2020 #4 Jomo said: You need to be complete. If u =e^x, then du =e^dx. This yields the integral posted by pka which is supposed to be known to you. Click to expand... aka \(\displaystyle du = e^x ~ dx\) -Dan

Jomo said: You need to be complete. If u =e^x, then du =e^dx. This yields the integral posted by pka which is supposed to be known to you. Click to expand... aka \(\displaystyle du = e^x ~ dx\) -Dan