# Find the area between two curves - Help finding the limits of integration - Again

## Homework Statement

The curves are:

$$y = \frac{x^{4}}{x^{2}+1}$$

and

$$y = \frac{1}{x^{2}+1}$$

## The Attempt at a Solution

So again I assume that:

$$\frac {x^{4}}{x^{2}+1} = \frac {1}{x^{2}+1}$$

and then cross multiply:

$$(x^{2}+1) = x^{4}(x^{2}+1)$$

not really sure at this point if i should distribute the x^4 but if i do it looks like so:

$$(x^{2}+1) = (x^{6}+x^{4})$$

so:

$$(x^{2}+1)-(x^{6}+x^{4}) = 0$$

and I am not really sure what to do at this point, I do have a polynomial if I do the subtraction which is:

$$-x^{6}-x^{4}+x^{2}+1 = 0$$

but I don't know how to factor it....

thanks guys!

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## Homework Statement

The curves are:

$$y = \frac{x^{4}}{x^{2}+1}$$

and

$$y = \frac{1}{x^{2}+1}$$

## The Attempt at a Solution

So again I assume that:

$$\frac {x^{4}}{x^{2}+1} = \frac {1}{x^{2}+1}$$

and then cross multiply:
Don't cross multiply. Multiply both sides by x^2 + 1.
$$(x^{2}+1) = x^{4}(x^{2}+1)$$

not really sure at this point if i should distribute the x^4 but if i do it looks like so:

$$(x^{2}+1) = (x^{6}+x^{4})$$

so:

$$(x^{2}+1)-(x^{6}+x^{4}) = 0$$

and I am not really sure what to do at this point, I do have a polynomial if I do the subtraction which is:

$$-x^{6}-x^{4}+x^{2}+1 = 0$$

but I don't know how to factor it....

thanks guys!

thanks man, worked out, but I am having a WAY hard time trying to do this problem so I will post the actual problem another thread. thanks for the help here though.