Integrate Sqrt[ e^x + 1 ]

  • Thread starter maxsthekat
  • Start date
  • #1
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Homework Statement



Integrate Sqrt[ ex+1]


2. The attempt at a solution
I first multiplied the equation by e^x / e^x, then tried substitution, with
u = e^x + 1
du = e^x dx

This gave me: Integral 1/(u-1) * (u)^1/2 du

However, I'm stumped at this part. Substitution here doesn't seem like it would work.

Any thoughts?

thanks!

-Max
 

Answers and Replies

  • #2
What if you let [itex] u=\sqrt{e^x+1} [/itex]??
 
  • #3
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What if you let [itex] u=\sqrt{e^x+1} [/itex]??

Then... du = [tex](1/2)*(e^x+1)^{-1/2} * e^x dx[/tex]

I don't see how that helps.
 
  • #4
35,287
7,140
I think your best shot might be a trig substitution. I haven't tried it, but that's what I would start with since the ordinary substitutions don't seem to lead anywhere and integration by parts also seems to be a dead end.

Trig substitutions are useful if you have a square root of the sum or difference of squares. In your case, ex = (ex/2)2.

Try tan u = ex/2 and see if that takes you anywhere.
 
  • #5
zcd
200
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If you let [tex]u=\sqrt{e^{x}+1}[/tex], then [tex]e^{x}\,dx=2u\,du[/tex] and [tex]\,dx=\frac{2u}{u^{2}-1}\,du[/tex]. This will leave you with [tex]\int\frac{2u^{2}}{u^{2}-1}\,du[/tex].
 
  • #6
i worked it out, checked with the help of Maple, and will help you out here to get what i got.

let [tex]u= e^{x}[/tex]
then, [tex] x= ln\left|u\right|[/tex]
and [tex]dx= \frac{du}{u}[/tex]

this should leave you with [tex] \frac{\sqrt{u+1}}{u}du[/tex]

and now consult your integration tables
 
  • #7
537
3
i worked it out, checked with the help of Maple, and will help you out here to get what i got.

let [tex]u= e^{x}[/tex]
then, [tex] x= ln\left|u\right|[/tex]
and [tex]dx= \frac{du}{u}[/tex]

this should leave you with [tex] \frac{\sqrt{u+1}}{u}du[/tex]

and now consult your integration tables

This substitution bothered me at first, because of the natural logarithm which gave rise to the absolute value sign. This could potentially limit what domain you're integrating over, but this doesn't matter since ex is always greater than zero, so you don't even need the absolute value around u. I think it's best to leave it out, because you ignored it anyways when you took the derivative.
 
  • #8
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Can this be done without having to resort to an integration table? I was told by the professor that we shouldn't need them for this problem.

Thanks, everyone, for all of your help :)
 
  • #9
Cyosis
Homework Helper
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All solvable integrals can be done without consulting an integration table. What have you tried so far? You have been given a few good suggestions, what have you done with them so far? I would suggest you continue where post #5 left off. If you get stuck show us how far you got and we'll help you from there.
 
  • #10
537
3
If you make this substitution, you can do it without integration tables.
[tex]
\begin{align*}
u &= \sqrt{e^x+1} \implies e^x = u^2-1 \\
du &= \frac{e^x}{2\sqrt{e^x+1}} \,dx \\
&= \frac{u^2-1}{2u} \,dx \\
\frac{2u}{u^2-1} \,du &= dx .
\end{align*}
[/tex]

Then
[tex]\int \sqrt{e^x+1} \,dx = \int \frac{2u^2}{u^2-1} \,du = \int \frac{2u^2}{(u-1)(u+1)}du[/tex]

Now expand this function by partial fractions and you will be able to integrate. I never like resorting to integration tables.

EDIT: I had missed post #5. It is the same substitution.
 
  • #11
1,101
3
I'm pretty sure the OP could have belt it out using the substitution in the original post. Note that if we let u = e^x + 1, then
[tex]
\int \sqrt{e^x+1} \,dx = \int \frac{\sqrt{u}}{u-1} \,du = \int \frac{\sqrt{u}}{(\sqrt{u}-1)(\sqrt{u}+1)}du = \int \frac{du}{u-1} + \int \frac{du}{\sqrt{u}+1}
[/tex]
*EDIT* Even though this is rather ugly.
 
  • #12
35,287
7,140
If you make this substitution, you can do it without integration tables.
[tex]
\begin{align*}
u &= \sqrt{e^x+1} \implies e^x = u^2-1 \\
du &= \frac{e^x}{2\sqrt{e^x+1}} \,dx \\
&= \frac{u^2-1}{2u} \,dx \\
\frac{2u}{u^2-1} \,du &= dx .
\end{align*}
[/tex]

Then
[tex]\int \sqrt{e^x+1} \,dx = \int \frac{2u^2}{u^2-1} \,du = \int \frac{2u^2}{(u-1)(u+1)}du[/tex]

Now expand this function by partial fractions and you will be able to integrate. I never like resorting to integration tables.

EDIT: I had missed post #5. It is the same substitution.
Before attempting to integrate that last expression, you should do long division, which gives you 2 + 2/(u^2 - 1). Then you can use partial fractions on the latter expression.
 
  • #13
537
3
Before attempting to integrate that last expression, you should do long division, which gives you 2 + 2/(u^2 - 1). Then you can use partial fractions on the latter expression.

Thanks for catching that.
 

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