Integrating √(1+x2)/x: A Step-by-Step Guide

[Saterial]

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.

 

[SammyS]

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

$\displaystyle \int \frac{\sqrt{1+x^2}}{x}\,dx \ ?$​

A better substitution would be a trig substitution such as x = tan(θ).

 

[Saterial]

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

 

[SammyS]

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 .

 

[Millennial]

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

 

[Saterial]

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?

 

[SammyS]

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.