# Intergation problem involving square root

## Homework Statement

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

## 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

## Answers and Replies

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gb7nash
Homework Helper

## Homework Statement

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

## 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)

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

Mark44
Mentor

## 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.

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
Homework Helper
$$\int_{1}^{3}\sqrt{1+\left(y^2-\frac{1}{4y^2}\right)^{2}}dy$$
$$\int_{1}^{3}\sqrt{1+y^2-\frac{1}{2}+\frac{1}{16y^4}}dy$$
The second line isn't right. Look again.

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