V val1 New member Joined Oct 17, 2005 Messages 40 Jun 13, 2007 #1 Hi. Can someone help with a couple of questions that I have? I am revising for an exam. Find an expression for the indefinite integral \(\displaystyle \L \int {\frac{{e^{2x} }}{{\sqrt {1 + e^{2x} } }}dx}\) Thank you. :?
Hi. Can someone help with a couple of questions that I have? I am revising for an exam. Find an expression for the indefinite integral \(\displaystyle \L \int {\frac{{e^{2x} }}{{\sqrt {1 + e^{2x} } }}dx}\) Thank you. :?
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Jun 13, 2007 #2 You could start by letting \(\displaystyle \L\\u=e^{2x}, \;\ du=2e^{2x}dx, \;\ \frac{du}{2}=e^{2x}dx\) Make the substitutions and you have: \(\displaystyle \L\\\frac{1}{2}\int\frac{1}{\sqrt{1+u}}du\) Now, can you finish?.
You could start by letting \(\displaystyle \L\\u=e^{2x}, \;\ du=2e^{2x}dx, \;\ \frac{du}{2}=e^{2x}dx\) Make the substitutions and you have: \(\displaystyle \L\\\frac{1}{2}\int\frac{1}{\sqrt{1+u}}du\) Now, can you finish?.
M morson Full Member Joined Apr 12, 2007 Messages 263 Jun 13, 2007 #3 \(\displaystyle u = 1 + e^{2x}\), \(\displaystyle du = 2e^{2x}dx\), \(\displaystyle \frac{du}{2}\ = e^{2x}dx\) = \(\displaystyle \L\ \frac{1}{2}\ \int u^{-1/2} \ du\) = \(\displaystyle \L\ sqrt{1 + e^{2x}} + C\)
\(\displaystyle u = 1 + e^{2x}\), \(\displaystyle du = 2e^{2x}dx\), \(\displaystyle \frac{du}{2}\ = e^{2x}dx\) = \(\displaystyle \L\ \frac{1}{2}\ \int u^{-1/2} \ du\) = \(\displaystyle \L\ sqrt{1 + e^{2x}} + C\)
