Length of a Curve: Find Length from 0 to 2

[skateza]

Homework Statement

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.

 

[Defennder]

Your question looks a lot like the one in the thread below, but only that 1/4(x+1) is the denominator of the whole expression. Were you guys using the same textbook?

https://www.physicsforums.com/showthread.php?t=210656

 

[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.