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Integral Help

  • Thread starter Saterial
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  • #1
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Homework Statement


∫√(1+x2)/x dx


Homework Equations





The Attempt at a Solution



Let u = 1+x2
du = 2xdx
1/2du=xdx
x=√(u-1)

∫√1+x2/x dx
=∫√u/√(u-1) du

or is it 1/2∫√udu as xdx would remove it. This is where I got confused.
 

Answers and Replies

  • #2
SammyS
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Homework Statement


∫√(1+x2)/x dx

Homework Equations



The Attempt at a Solution



Let u = 1+x2
du = 2xdx
1/2du=xdx
x=√(u-1)

∫√1+x2/x dx
=∫√u/√(u-1) du

or is it 1/2∫√udu as xdx would remove it. This is where I got confused.
What did you do with the x in the denominator of
[itex]\displaystyle \int \frac{\sqrt{1+x^2}}{x}\,dx \ ?[/itex]​

A better substitution would be a trig substitution such as x = tan(θ).
 
  • #3
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In my solution attempt, I tried two different ways. In your specific question, I replaced the x in the denominator with √(u-1) , as solving for x in u=1+x^2
 
  • #4
SammyS
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In my solution attempt, I tried two different ways. In your specific question, I replaced the x in the denominator with √(u-1) , as solving for x in u=1+x^2
You're missing an x.

You have du = 2x dx . Therefore dx = 1/(2x) du.

This gives you an x2 in the denominator. x2 = u - 1 .
 
  • #5
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A trigonometric substitution does the trick here, an hyperbolic substitution might also work.
 
  • #6
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Thanks for the help so far. With that little fix, I've now reached the point of

1/2 ∫√u/(u-1) du

I don't know where to go from here. I also am having many people tell me that this method will not work? That Trig-Substitution is the only way to solve this problem... Is that true? Am I just wasting my time?
 
  • #7
SammyS
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Thanks for the help so far. With that little fix, I've now reached the point of

1/2 ∫√u/(u-1) du

I don't know where to go from here. I also am having many people tell me that this method will not work? That Trig-Substitution is the only way to solve this problem... Is that true? Am I just wasting my time?
There are two ways that I know of to do this integration.

Use either a trig substitution, such as x = tan(θ) , or use a substitution involving hyperbolic functions, such as x = sinh(u).

These are handy because, tan2(θ) + 1 = sec2(θ) and sinh2(u) + 1 = cosh2(u).


Either substitution will require you to do a far amount of follow-up work to finish the integration.
 

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