How do I express x in terms of y for the following expression ? $y=2x+\frac{8}{x^2}-5$ How do I express x in terms of y for the following expression ?
$$y=2x+\frac{8}{x^2}-5$$
The reason I want to do this is to calculate the area between this curve, the y-axis, y=5 and y=1.
I found the answer to be 5 using a roundabout method.
But my first attempt was to try and express x in terms of y. 
Basic algebra gave me an expression with an $x^3$.
 A: Basic algebra is correct.
From
$y=2x+\frac{8}{x^2}-5
$,
multiplying by $x^2$
we get
$yx^2=2x^3+8-5x^2
$,
or
$2x^3-(5-y)x^2+8
= 0
$.
This is,
unfortunately,
a cubic
which can be solved by the
traditionally messy formula.
Its derivative is
$6x^2-2(5-y)x
= 0
$
which has roots
$x=0$
and
$x=\dfrac{5-y}{3}
$.
Note that
at $x=0$
the function is $8$
and at
$\dfrac{5-y}{3}
$
the value is
$2(\frac{5-y}{3})^3-(5-y)(\frac{5-y}{3})^2+8
=(\frac{5-y}{3})^3(2-3)+8
=-(\frac{5-y}{3})^3+8
=(\frac{y-5}{3})^3+8
$.
From this,
we can determine which
values of $y$
give an equation with
$1$ or $3$ real roots.
A: To find the area, there isn't a particular reason to find $x$ in terms of $y$. A graph of the function shows that the area is relatively simple to find.

You just have to find the area under your function from $1$ to $2$ (you can use the Fundamental Theorem of Calculus here), and manipulate it to get your desired area.
A: I think that this will get really messy if you integrate with respect to y.
Here is what I suggest you do.
$y = 2x + \frac 8{x^2} -5$
Evaluated at $y = 1$ and $y = 5$
$1 = 2x + \frac 8{x^2} -5\\
x = 2$
$5 = 2x + \frac 8{x^2} -5\\
x = 1$
$\int_0^1 (5-1) \;dx + \int_1^2  2x + \frac 8{x^2} -5 - 1\;dx$
