How can I simplify this derivative to make calculating arc length easier?

[jen333]

Hi! Here's my question on finding arc length. If I've taken the derivative correctly, is there anyway I can simplify it before putting it into the arc length formula?

Homework Statement

Find the arc length where 0\leqx\leq2
y=(x^{3}/3)+x^{2}+x+1/(4x+4)

Homework Equations

L=\intds=\sqrt{1+(dy/dx)^{2}}

The Attempt at a Solution

I've only taken the derivative so far:
(dy/dx)=x^{2}+2x+1-4(4x+4)^{-2}
=(x+1)^{2}-4(4x+4)^{-2}


I tried expanding the equation, but that only makes it more complex.

I know to find the arc length I need to square the derivative and place it in the formula (and possibly using substitution), but I'm just wondering how I can simplify the above equation to make it easier to square and calculate!

 

[Gib Z]

Polynomial division! =]

 

[Defennder]

Yeah you should get this if you do the polynomial division correctly:

\int_0^2 (x+1)^2 + \frac{1}{4(x+1)^4} dx.

To make things easier, denote (x+1) by A or some other letter, then it'll look easier.

EDIT: Fixed error

 

[HallsofIvy]

I would look at it like this:
y'= x^2+ 2x+ 1- \frac{4}{(4x+ 4)^2}= x^2+ 2x+ 1- \frac{1}{4(x+1)^2}
= (x+ 1)^2- \left(\frac{1}{2(x+1)}\right)^2
so
(y')^2= (x+1)^4- 1/2+ \left(\frac{1}{2(x+1)}\right)^4
(y')^2+ 1= (x+1)^4+ 1/2+ \left(\frac{1}{2(x+1)}\right)^4
= \left((x+1)^2+ \left(\frac{1}{2(x+1)}\right)^2\right)^2
so that
\sqrt{(y')^2+ 1}= (x+1)^2+ \frac{1}{2(x+2)}

 

[Defennder]

= \left((x+1)^2+ \left(\frac{1}{2(x+1)}\right)^2\right)^2
so that
\sqrt{(y')^2+ 1}= (x+1)^2+ \frac{1}{2(x+2)}

There's a small error here, the last term should read 1/4(x+1)^2 instead.