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## Homework Statement

So the curves are:

[tex] y = \frac {x^{4}}{x^{2}+1} [/tex]

and

[tex] y = \frac {1}{x^{2}+1} [/tex]

## The Attempt at a Solution

So first the limits of integration are:

[tex] \frac{x^{4}}{x^{2}+1} = \frac{1}{x^{2}+1} [/tex]

[tex] (x^{2}+1)\frac{x^{4}}{x^{2}+1} = \frac{1}{x^{2}+1} (x^{2}+1)[/tex]

[tex] x^{4} = 1 [/tex]

[tex] x^{4}-1 = 0 [/tex]

[tex] (x^{2}+1)(x^{2}-1)= 0 [/tex]

so

[tex] x^{2}+1 = 0 \rightarrow x^{2} =-1[/tex]

and since:

[tex] \sqrt{-1}=i[/tex]

I assume that it should be the negative square root of 1:

[tex] x= -\sqrt{1} = -1 [/tex]

and also:

[tex] x^{2}-1 = 0 \rightarrow x^{2}=1 \rightarrow x= \sqrt{1}=1[/tex]

so the integral should go from -1 to 1

Also I can tell from the graph of the two functions that:

[tex] y = \frac {1}{x^{2}+1} [/tex]

bounds

[tex] y = \frac {x^{4}}{x^{2}+1} [/tex]

so the integral should be set up this way (I will use two integrals purely for clarity):

[tex]\int^{1}_{-1} (x^{2}+1)^{-1} dx - \int^{1}_{-1} \frac{x^{4}}{x^{2}+1} dx [/tex]

I am fairly certain that the integral on the left can be evaluated using this anti-derivative:

[tex] \frac{1}{2x}ln(x^{2}+1) [/tex]

which I got by somewhat trial and error as u substitution did me no good on that problem, but that's easy one, I can not figure out the right side for the life of me, here's my shot at it, but its not very good.

Let:

[tex] u = x^{2}+1 [/tex]

[tex] du = 2xdx [/tex]

[tex] dx = \frac {1}{2}x^{-1} [/tex]

so the problem now looks like:

[tex] \int^{1}_{-1} \frac{1}{2}\frac{x^{4}}{(u)(x)} du [/tex]

and the x will remove 1 exponent so:

[tex] \int^{1}_{-1} \frac{1}{2}\frac{x^{3}}{(u)} du [/tex]

which can be put:

[tex] \int^{1}_{-1} \frac{1}{2}(x)^{3}(u)^{-1} du [/tex]

and since

[tex] u = x^{2}+1 \rightarrow x = \sqrt{u-1} [/tex]

so now its:

[tex] \int^{1}_{-1} \frac{1}{2}(\sqrt{u-1})^{3}(u)^{-1} du [/tex]

and

[tex] (\sqrt{u-1})^{3} = (u-1)\sqrt{u-1} = u\sqrt{u-1} -\sqrt{u-1} [/tex]

and put together:

[tex] \int^{1}_{-1} \frac{1}{2}\frac{u\sqrt{u-1} -\sqrt{u-1}}{u} du [/tex]

which cancels 1 u to make it:

[tex] \int^{1}_{-1} \frac{1}{2}\sqrt{u-1} -\frac {\sqrt{u-1}}{u} du [/tex]

and I am just not seeing this work out from here.... did I make this too hard or what?

I guess I should say that I have tried u = x^4 as well, and it does work out slightly better but I get stuck at the point of:

[tex] \int^{1}_{-1} \frac{1}{4} \frac{u}{(u+u^{3/4})} [/tex]

If i go that way I can't figure out how to do:

[tex] u * (u+u^{3/4})^{-1} [/tex]

if it wasn't the reciprocal of u+u^3/4 it would go....

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