- #1

- 54

- 0

## 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.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Saterial
- Start date

- #1

- 54

- 0

∫√(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

Staff Emeritus

Science Advisor

Homework Helper

Gold Member

- 11,368

- 1,035

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

- 54

- 0

- #4

SammyS

Staff Emeritus

Science Advisor

Homework Helper

Gold Member

- 11,368

- 1,035

You're missing an x.

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

This gives you an x

- #5

- 296

- 0

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

- #6

- 54

- 0

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

Staff Emeritus

Science Advisor

Homework Helper

Gold Member

- 11,368

- 1,035

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.

Share: