# Length of a curve

1. Jan 24, 2008

### skateza

1. The problem statement, all variables and given/known data

I need to find the length of $$\frac{\frac{1}{3}x^{3} + x^{2} + x + 1}{4x+4}$$ from x=0 to x=2 but i can not factor this down to be able to set up the integral, any suggestions, here is the derivative:
$$\frac{\frac{8}{3}x^{3}+8x^{2}+8x}{16x^{2}+32x+16}$$
It is possible I might have to factor the derivative rather than the function itself which is why i supplied both.

2. Jan 25, 2008

### Defennder

3. Jan 25, 2008

### HallsofIvy

I would write this as:
$$\frac{1}{12} \frac{(x^3+ 3x^2+ 3x+ 1)+ 2}{x+1}=\frac{1}{12}\frac{(x+1)^3+ 2}{x+1}= \frac{1}{12}((x+1)^2+ \frac{2}{x+1})$$
That looks to me like it will be easier to handle.