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