Intergation problem involving square root

[miglo]

Homework Statement

$$\int_{1}^{3}\sqrt{y^2+\frac{1}{2}+\frac{1}{16y^4}}dy$$

Homework Equations

The Attempt at a Solution

I don't know what to do with this integral, but i do feel like this can be simplified into something much simpler.
I tried rewriting it as $\frac{1}{4}\int_{1}^{3}\sqrt{\frac{16y^6+8y^4+1}{y^4}}dy$
but I'm still not getting the same answer as in the back of the book

 

[gb7nash]

Homework Statement

$$\int_{1}^{3}\sqrt{y^2+\frac{1}{2}+\frac{1}{16y^4}}dy$$

Homework Equations

The Attempt at a Solution

I don't know what to do with this integral, but i do feel like this can be simplified into something much simpler.

It can. Fill in the blanks.

$y^2+\frac{1}{2}+\frac{1}{16y^4}$ = (_ + _)²

edit: I take it back. I thought it was 1/(16y²)

 

[miglo]

Is it $$\left(y^2+\frac{1}{4y^2}\right)^{2}$$?

 

[Mark44]

Homework Statement

$$\int_{1}^{3}\sqrt{y^2+\frac{1}{2}+\frac{1}{16y^4}}dy$$

Is it $$\left(y^2+\frac{1}{4y^2}\right)^{2}$$?

No, given the original problem.

Are you sure that you copied it correctly?

Could the problem have been this integral?

$$\int_{1}^{3}\sqrt{y^4+\frac{1}{2}+\frac{1}{16y^4}}dy$$

If that was the problem, the part inside the radical factors nicely.

 

[miglo]

yeah that doesn't work.
well its an arc length problem
$$x=\frac{y^3}{3}+\frac{1}{4y}$$
$$\frac{dx}{dy}=y^2-\frac{1}{4y^2}$$
$$\int_{1}^{3}\sqrt{1+\left(y^2-\frac{1}{4y^2}\right)^{2}}dy$$
$$\int_{1}^{3}\sqrt{1+y^2-\frac{1}{2}+\frac{1}{16y^4}}dy$$
$$\int_{1}^{3}\sqrt{y^2+\frac{1}{2}+\frac{1}{16y^4}}dy$$

 

[gb7nash]

$$\int_{1}^{3}\sqrt{1+\left(y^2-\frac{1}{4y^2}\right)^{2}}dy$$
$$\int_{1}^{3}\sqrt{1+y^2-\frac{1}{2}+\frac{1}{16y^4}}dy$$

The second line isn't right. Look again.

 

[miglo]

Yeah I see what I'm doing wrong.
It's supposed to be $y^4$
thanks mark44 and gb7nash