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!

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.

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?

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.

[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, 200916 yr by buttacup
Consecutive posts merged.

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, 200916 yr by D H
Fix math error

I can sleep better now............... thank you!

Wolframalpha.com is a really good site if you need help with a math problem