Find Length of Curve: y=x^2+2 from 0-3

[tandoorichicken]

Find the length of $$y = \frac{1}{3} (x^2 + 2)^{\frac{3}{2}}$$ from x=0 to x=3.

I used the formula $$s = \int_{a}^{b} \sqrt{1 + \frac{\,dy}{\,dx}} \,dx$$.
After plugging everything in, I got
$$s = \int_{0}^{3} (1 + \frac{1}{4} (x^2 + 2))}^\frac{1}{2} \,dx$$
Now, this isn't an integral I've learned how to do, so
1) Did I do anything wrong, and
2) If yes, what?

 

[tandoorichicken]

I realized I forgot to use the chain rule when doing dy/dx. However when I do that, I get even a messier integral,
$$s = \int_{0}^{3} (1 + x^2(x^2 + 2))^\frac{1}{2} \,dx$$

?

 

[himanshu121]

It should be

$$s = \int_{a}^{b} \sqrt{1 + (\frac{\,dy}{\,dx})^2} \,dx$$

But that's not the pro u just wrote the wrong one and applied the right

u also have
$$1+x^2(x^2+2) = 1+x^4+2x^2=x^4+2x^2+1=(x^2+1)^2$$
Is it enough

 

[tandoorichicken]

Woah yeah, thanks, overlooked a simple thing there.