- #1

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## Homework Statement

∫√(1+x

^{2})/x dx

## Homework Equations

## The Attempt at a Solution

Let u = 1+x

^{2}

du = 2xdx

1/2du=xdx

x=√(u-1)

∫√1+x

^{2}/x dx

=∫√u/√(u-1) du

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

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- Thread starter Saterial
- Start date

- #1

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∫√(1+x

Let u = 1+x

du = 2xdx

1/2du=xdx

x=√(u-1)

∫√1+x

=∫√u/√(u-1) du

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

- #2

SammyS

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Homework Helper

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What did you do with the x in the denominator of## Homework Statement

∫√(1+x^{2})/x dx

## Homework Equations

## The Attempt at a Solution

Let u = 1+x^{2}

du = 2xdx

1/2du=xdx

x=√(u-1)

∫√1+x^{2}/x dx

=∫√u/√(u-1) du

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

[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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- #4

SammyS

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You're missing an x.

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

This gives you an x

- #5

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A trigonometric substitution does the trick here, an hyperbolic substitution might also work.

- #6

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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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There are two ways that I know of to do this integration.

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?

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, tan

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

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