Integral involving square root and exp

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  • #1
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Homework Statement



[itex]\int[/itex][itex]\frac{dx}{\sqrt{e^{x} + 1}}[/itex]

Homework Equations


Using u-substitution


The Attempt at a Solution



Let u = [itex]\sqrt{e^{x} + 1}[/itex] [itex]\Rightarrow[/itex] u[itex]^{2} - 1[/itex] = e[itex]^{x}[/itex]
Then, du = [itex]\frac{e^{x} dx}{2\sqrt{e^{x} + 1}}[/itex] [itex]\Rightarrow[/itex] dx = [itex]\frac{2u du}{u^{2}-1}[/itex]

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

But, I'm stuck at this point. I think I want to break it up into two simpler integrals, but I'm not sure how to do this. Any suggestions would be greatly appreciated!
 

Answers and Replies

  • #2
Dick
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Homework Statement



[itex]\int[/itex][itex]\frac{dx}{\sqrt{e^{x} + 1}}[/itex]

Homework Equations


Using u-substitution


The Attempt at a Solution



Let u = [itex]\sqrt{e^{x} + 1}[/itex] [itex]\Rightarrow[/itex] u[itex]^{2} - 1[/itex] = e[itex]^{x}[/itex]
Then, du = [itex]\frac{e^{x} dx}{2\sqrt{e^{x} + 1}}[/itex] [itex]\Rightarrow[/itex] dx = [itex]\frac{2u du}{u^{2}-1}[/itex]

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

But, I'm stuck at this point. I think I want to break it up into two simpler integrals, but I'm not sure how to do this. Any suggestions would be greatly appreciated!
Partial fractions is what you want. Factor the denominator.
 
  • #3
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Okay.

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

And now I'm stuck again.
 
  • #4
Dick
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Okay.

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

And now I'm stuck again.
Partial fractions! ##\frac{1}{(u+1)(u-1)}=\frac{A}{u+1}+\frac{B}{u-1}## for some constants A and B. Find those constants.
 
  • #5
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Okay. I think I've got it.

So, [itex]\int[/itex][itex]\frac{dx}{\sqrt{e^{x} + 1}}[/itex] = [itex]\int[/itex][itex]\frac{2u du}{u(u^{2}-1)}[/itex] = [itex]\int[/itex][itex]\frac{2du}{(u+1)(u-1)}[/itex] = [itex]\int[/itex][itex]\frac{du}{u-1}[/itex] - [itex]\int[/itex][itex]\frac{du}{u-1}[/itex] = ln|u-1| - ln|u+1| = ln[itex]\frac{|u-1|}{|u+1|}[/itex] where u = [itex]\sqrt{e^{x}+1}[/itex]

= ln[itex]\frac{\sqrt{e^{x}+1}-1}{\sqrt{e^{x}+1}+1}[/itex]

Thank you for all your help.
 

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