# Arc length: Can't Solve the Integral

1. Feb 21, 2010

### swimgirl5892

Find the exact length of the curve analytically by antidifferentiation:
y = (x3/3) + x2 + x + (1/(4x +4)) on the interval 0 < x < 2

So I set it up using the length of a curve formula:

L = $$\int\sqrt{1+(x^2+2x+1+(\frac{-1}{4(x+1)^2}}$$

And simplified it to
L = $$\int\sqrt{\frac{1}{2}+(x+1)^4+\frac{1}{16(x+1)^4}}$$

But I cannot figure out how to antidifferentiate this! Any help is appreciated :)

2. Feb 21, 2010

### Dustinsfl

If you establish a LCD, you obtain another polynomial to the 8th degree. With this polynomial you can call the smaller degree w, then all you need to do is factor. From this, you will have something squared divide by the denominator.

3. Feb 21, 2010

### swimgirl5892

Ok, so then when I find an LCD I get:

$$\int\sqrt{\frac{8(x+1)^4}{16(x+1)^4}+\frac{16(x+1)^8}{16(x+1)^4}+\frac{1}{16(x+1)^4}}$$

which simplifies to

$$\int\frac{4(x+1)^4 +1}{4(x+1)^2}$$

Is this as simple as possible? I'm afraid I still can't integrate this...

4. Feb 21, 2010

### Dustinsfl

For example, (4+5)/1 = 4/1 + 5/1. Does that help?

5. Feb 21, 2010

### kbaumen

Doesn't u=x+1 help?

6. Feb 21, 2010

### swimgirl5892

I got it!!!! The answer is $$\frac{53}{6}$$.

Thank you!!!

7. Feb 21, 2010

### Dustinsfl

That is what I obtained too.