Veritas Posted May 30, 2009 Share Posted May 30, 2009 Hi everyone! This is my first post on this site so if I'm breaking any rules I appolgize in advance. For an opcoming physics entrance exam i'm repolishing my math skills but I just can't seem to solve this problem: Integrate: (e^x)/(1+e^(3x)) dx I've tried substituting with : u = e^x, u = e^3x and u= 1+e^3x, but nothing seems to work out. Maybe i'm just missing something very obvious but I'm just lost at this point thanks in advance! Link to comment Share on other sites More sharing options...
D H Posted May 30, 2009 Share Posted May 30, 2009 The u-substitution u=e^x is a good place to start. Then du=e^x dx, so [math]\int\frac{e^x}{1+e^{3x}} dx\quad\to\quad\int \frac {du}{u^3+1}[/math] So this changes the problem from an exponential integral to a rational polynomial. Next step: Use the fact that u+1 divides any polynomial of the form un+1. You will get u+1 times a quadratic. Use partial fractions. You'll still have an ugly beast, but a completely solvable ugly beast. Link to comment Share on other sites More sharing options...
Veritas Posted May 31, 2009 Author Share Posted May 31, 2009 Thanks for your quick response, but don't quite understand what you mean by "Use the fact that u+1 divides any polynomial of the form un+1". I'm from Holland so maybe you're using different terminology than I'm used to but could you please explain me that part a bit more? Link to comment Share on other sites More sharing options...
D H Posted May 31, 2009 Share Posted May 31, 2009 Divides, in that context, is a synonym for "is a divisor of". 2 and 3 divide 24, but 5 does not. I was a bit hasty in my typing. I said that u+1 divides any polynomial of the form un+1. I should have qualified that with if n is odd. Fortunately, 3 is odd. In other words, there exists a quadratic q(x) such that (u+1)*q(x) = u3+1. Link to comment Share on other sites More sharing options...
Xittenn Posted June 13, 2009 Share Posted June 13, 2009 (edited) [math]\int\frac{e^x}{e^x+1}\frac{1}{e^{2x}-e^x+1}dx [/math] [math]\int\frac{4e^x}{3(e^x+1)}\frac{1}{(\frac{e^{2x}-e^x+1}{3})+1}dx [/math] [math]\frac{4}{3}\int\frac{e^x}{(e^x+1)}\frac{1}{(\frac{e^{2x}-e^x+1}{3})+1}dx [/math] [math] x' = \frac{-1+2e^x}{\sqrt3}[/math] used prime to denote other x [math] dv = \frac{e^x}{1+x'^2}dx = \frac{1}{1+x'^2}dx' [/math] [math] v = tan^{-1} [\frac{-1+2e^x}{\sqrt3}][/math] [math] u = \frac{1}{1+e^x} [/math] [math] du = -\frac{e^x}{(1+e^x)^2} [/math] [math] \frac{2\sqrt{3}}{3(e^x+1)}tan^{-1}[\frac{-1+2e^x}{\sqrt{3}}]+\frac{2\sqrt{3}}{3}\int\frac{e^x}{(e^x+1)^2}tan^{-1}[\frac{-1+2e^x}{\sqrt3}]dx [/math] ummmmmmmmmmmm 1) [math] \int tan^{-1} \frac{x}{a} dx = x [tan^{-1} [\frac{x}{a}] - \frac{a}{2}ln(a^2 + x^2)] [/math] those poor trees................ 2) [math] \frac {\sqrt{3}e^x}{2-2e^x+2e^{2x}}dx [/math] better....... [math] u=tan^{-1}[\frac{-1+2e^x}{\sqrt3}] [/math] [math] du=\frac {\sqrt{3}e^x}{2-2e^x+2e^{2x}}dx [/math] [math] dv=\frac{e^x}{(1+e^x)^2}dx[/math] [math] v = - \frac {1}{1+e^x} [/math] [math] \frac{2\sqrt{3}}{3}(\frac{1}{1+e^x}tan^{-1}[\frac{-1+2e^x}{\sqrt{3}}]-\frac {1}{1+e^x}tan^{-1}[\frac{-1+2e^x}{\sqrt3}]+\int\frac {1}{1+e^x}\frac {\sqrt{3}e^x}{2-2e^x+2e^{2x}}dx) [/math] [math] \frac{2\sqrt{3}}{3}(\int\frac {1}{1+e^x}\frac {\sqrt{3}e^x}{2-2e^x+2e^{2x}}dx) [/math] [math] \int\frac {1}{1+e^x}\frac {e^x}{1-e^x+e^{2x}}dx [/math] ooooopssy [math] \int\frac {e^{-x}}{1+e^{-x}}\frac {e^x}{1-e^x+e^{2x}}dx [/math] [math] u = \frac {e^x}{1-e^x+e^{2x}} [/math] [math] dv = \frac {e^{-x}}{1+e^{-x}} dx [/math] Wolfram result: [math] \frac{tan^{-1} [\frac{-1+2e^x}{\sqrt3}]}{\sqrt3}+\frac{1}{3}ln(1+e^x)-\frac{1}{6}ln(1-e^x+e^{2x})[/math] Edited June 13, 2009 by buttacup Consecutive posts merged. Link to comment Share on other sites More sharing options...
D H Posted June 13, 2009 Share Posted June 13, 2009 (edited) You didn't use the hint to make the u-substitution. When you do make a u-substitution, don't be so quick to back-substitute. From post #2, The u-substitution u=e^x is a good place to start. Then du=e^x dx, so [math]\int\frac{e^x}{1+e^{3x}} dx\quad\to\quad\int \frac {du}{u^3+1}[/math] Now factor [math]u^3+1[/math] as [math]u^3+1=(u+1)(u^2-u+1)[/math] and use partial fractions: [math] \frac 1 {u^3+1} = \frac 1 3 \left(\frac 1 {u+1} - \frac{u-2}{u^2-u+1}\right) [/math] Note that [math]d(u^2-u+1) = 2u-1[/math]. That second factor is a bit easier to deal with if it is rewritten using this fact: [math]\frac{u-2}{u^2-u+1} = \frac 1 2 \left(\frac{2u-1}{u^2-u+1} \frac{3}{u^2-u+1}\right)[/math] With this, [math] \frac 1 {u^3+1} = \frac 1 3 \,\frac 1 {u+1} - \frac 1 6\, \frac{2u-1}{u^2-u+1} + \frac 1 2\,\frac 1 {u^2-u+1} [/math] Now integrate: [math] \int \frac{du}{u^3+1} = \frac 1 3 \ln(u+1) - \frac 1 6 \ln(u^2-u+1) +\frac 1 2 \int \frac{du}{u^2-u+1} [/math] How to deal with this final integral? Make another u-substitution. Write [math]u=av+b[/math] such that [math]u^2-u+1=c(v^2+1)[/math]. The solution is [math]u=\frac 1 2 \left(\sqrt 3 v+1\right)[/math]. With this, [math]du=\frac{\sqrt 3}{2} dv[/math] and [math]u^2-u+1=c(v^2+1) = \frac 3 4 (v^2+1)[/math]. Thus [math]\int \frac{du}{u^2-u+1} = \frac 2 {\sqrt 3} \int \frac{dv} {v^2+1} = \frac 2 {\sqrt 3}\tan^{-1} v[/math] Now back-substitute with [math]v=\frac{2u-1}{\sqrt 3}[/math]: [math]\int \frac{du}{u^2-u+1} = \frac 2 {\sqrt 3}\tan^{-1}\left(\frac{2u-1}{\sqrt 3}\right)[/math] And thus [math] \int \frac{du}{u^3+1} = \frac 1 3 \ln(u+1) - \frac 1 6 \ln(u^2-u+1) +\frac 1 {\sqrt 3} \tan^{-1}\left(\frac{2u-1}{\sqrt 3}\right) [/math] Back-substitute again, with [math]u=e^x[/math]: [math] \int\frac{e^x}{1+e^{3x}} dx = \frac 1 3 \ln(e^x+1) - \frac 1 6 \ln(e^{2x}-e^x+1) +\frac 1 {\sqrt 3} \tan^{-1}\left(\frac{2e^x-1}{\sqrt 3}\right) [/math] Edited June 14, 2009 by D H Fix math error Link to comment Share on other sites More sharing options...
Xittenn Posted June 13, 2009 Share Posted June 13, 2009 I can sleep better now............... thank you! Link to comment Share on other sites More sharing options...
KtownChemist Posted June 18, 2009 Share Posted June 18, 2009 Wolframalpha.com is a really good site if you need help with a math problem Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now