How Can I Calculate Arc Length Without a Calculator?

[lastdayx52]

How in gods name do I do that? I attempted that integral and... it just can't be integrated!

What I tried:

That doesn't help one bit... How do I do this? NOTE: No graphing calculator is to be used.

 

[Mark44]

You have as the integrand
$$\sqrt{1 + (x - 1/(4x))^2}$$
$$= \sqrt{1 + (x^2 - 1/2 + 1/16x^2)}$$

When you group together the only terms that can be grouped, you'll have a perfect square under the radical.

 

[latentcorpse]

when you squared out the (x-x/4)^2 you should have got 1/2 not 3/2. this should simplify things hopefully cause you should be able to factorise the numerator into
sqrt[(4x^2+1)^2]

 

[lastdayx52]

You have as the integrand
$$\sqrt{1 + (x - 1/(4x))^2}$$
$$= \sqrt{1 + (x^2 - 1/2 + 1/16x^2)}$$

When you group together the only terms that can be grouped, you'll have a perfect square under the radical.

Yes that gives me:

However, what good is that, since there's still a 1 in there? I can't squareroot it to simplify...

when you squared out the (x-x/4)^2 you should have got 1/2 not 3/2. this should simplify things hopefully cause you should be able to factorise the numerator into
sqrt[(4x^2+1)^2]

You do get 1/2, but I added a 1, therefore 3/2.

 

[Mark44]

You have as the integrand
$$\sqrt{1 + (x - 1/(4x))^2}$$
$$= \sqrt{1 + (x^2 - 1/2 + 1/16x^2)}$$

When you group together the only terms that can be grouped, you'll have a perfect square under the radical.

Continuing from this point...
$$= \sqrt{x^2 + 1/2 + 1/16x^2}$$
The part under the radical is a perfect square. Surely you can take it from here!

 

[lastdayx52]

OH... I added wrong... wow... Stupid mistakes FTL... Thanks all!