How would you find the anti derivative of (1+x^2)^(1/2)

[skyturnred]

Homework Statement

How would you find the anti derivative of (1+x^2)^(1/2)

I am trying to find the definite integral of that function from 0 to 1

Homework Equations

The Attempt at a Solution

Here is what I tried so far: I let u=1+x^2
du=2xdx
solving for dx, dx=(1/2x)du
I plug dx into the integral to get
(integral sign)(1/(2x))(u)^(1/2)du
So after I solve this I get [(1+x^2)^(3/2)]/(3x) from 0 to 1. But obviously this is wrong because if I plug 0 in for x, the function does not exist!

 

[Stimpon]

I'm not going to tell you which one but using a trigonometric identity is probably the simplest way.

 

[vela]

Homework Statement

How would you find the anti derivative of (1+x^2)^(1/2)

I am trying to find the definite integral of that function from 0 to 1

Homework Equations

The Attempt at a Solution

Here is what I tried so far: I let u=1+x^2
du=2xdx
solving for dx, dx=(1/2x)du
I plug dx into the integral to get
(integral sign)(1/(2x))(u)^(1/2)du
So after I solve this I get [(1+x^2)^(3/2)]/(3x) from 0 to 1. But obviously this is wrong because if I plug 0 in for x, the function does not exist!

You can't simply treat the remaining x as a constant. You have to write everything in terms of u. If you solve for x in terms of u, you get $x=\sqrt{u-1}$, so you end up with
$$\int\frac{1}{2x}u^{1/2}\,du = \int\frac{1}{2\sqrt{u-1}}u^{1/2}\,du$$This doesn't look much better than what you started with, so you'll want to try something different.

When you see something like 1+x², you should think "trig substitution."

 

[skyturnred]

You can't simply treat the remaining x as a constant. You have to write everything in terms of u. If you solve for x in terms of u, you get $x=\sqrt{u-1}$, so you end up with
$$\int\frac{1}{2x}u^{1/2}\,du = \int\frac{1}{2\sqrt{u-1}}u^{1/2}\,du$$This doesn't look much better than what you started with, so you'll want to try something different.

When you see something like 1+x², you should think "trig substitution."

OK thanks! We haven't even learned integration by parts yet, let alone trig substitution. So I must not be able to solve this problem.