Arc length: Can't Solve the Integral

[swimgirl5892]

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

 

[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.

 

[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...

 

[Dustinsfl]

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

 

[kbaumen]

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...

Doesn't u=x+1 help?

 

[swimgirl5892]

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


Thank you!

 

[Dustinsfl]

That is what I obtained too.