No messy calculations, please!

Note that $(y-x)( (y-x) + 2x)^2 = (y-x)(y+x)^2 = y^3 + y^2x - x^2y - x^3 = 1$ so if we set $z = y-x,$ then $z(z+2x)^2 = 1 \implies x = \frac{1}{2} \left(z^{-1/2} - z\right) \implies dx = \frac{1}{2}\left(-\frac{1}{2}z^{-3/2} - 1\right)\,dz$ We can now use substitution to evaluate $\begin{aligned} \int_0^\infty (y-x)\,dx &= \int_1^0 z\cdot \frac{1}{2}\left(-\frac{1}{2}z^{-3/2} - 1\right)\,dz \\ &= \int_0^1 \left(\frac{1}{4}z^{-1/2} + \frac{1}{2}z\right)\,dz \\ &= \left. \frac{1}{2}z^{1/2} + \frac{1}{4}z^2 \right|_0^1 \\ &= \frac{1}{2} + \frac{1}{4} \\ &= \frac{3}{4} \end{aligned}$

By factorisation, we can conclude that: $y-x = \frac{1}{(y+x)^2}$

Now, this could have something to do with rotations. The intuition for this should arise from the fact that while rotating a curve by 45 degrees using a rotation matrix , we usually replace x and y with either the sum or the difference of x and y divided by the square root of 2. Also, simply sketching the function might have helped.

Hence:

$\sqrt{2}\frac{(y-x)}{\sqrt{2}} = \frac{1}{2(\frac{y+x}{\sqrt{2}})^2}$

Let's now try to describe the given curve in a coordinate system rotated by 45 degrees counterclockwise. The new coordinates are given as:

$\begin{pmatrix}x' \\ y' \end{pmatrix} = \begin{pmatrix} \frac{\sqrt2}{2} &\frac{\sqrt2}{2} \\ -\frac{\sqrt2}{2}& \frac{\sqrt2}{2} \end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}$

And so, we can happily conclude that:

$y' = \frac{1}{2\sqrt{2}x'^2}$

Since $y = x$ is the asymptote of our function (it is simply $y= 0$ rotated by 45 degrees counterclockwise), it is geometrically simple to conclude that our integral is equal to (if needed, I will supply a drawing when I have more time):

$\frac{1}{4} + \frac{1}{2\sqrt2} \int_{\frac{\sqrt2}{2}}^{\infty} \frac{1}{x^2}dx$ which gives the anwser, $\frac{3}{4}$ .

Use the rotation transform to turn this implicit function by $45$ degrees to the following explicit function of $x$ only

$y=\dfrac{1}{2\sqrt{2}x^2}$

and integrate $y=x$ from $x=0$ to $\dfrac{1}{\sqrt{2}}$ , and the other from $\dfrac {1}{\sqrt{2}}$ to infinity.

Factorisation gives (y-x) = 1/(x+y)^2. Now, introduce u = x+y. This will give (y-x) = u - 2x. Plug this substitution into the above equality to find x as an explicit function of u.

You will get x = (u^3 - 1)/(2*u^2). Using this, find dx in terms of du.

You will get dx = du (u^3 + 2)/(2 u^3).

Originally we had to integrate (y - x) wrt x. This is equivalent to integrating 1/(x+y)^2 wrt x due to the factorisation process. This is also equivalent to integrating dx/u^2 with dx replaced by the above obtained equality which contains du.

Finally, the integral becomes (u^3 + 2)/(2*u^5) wrt u.

The limits are slightly tricky. But with a little bit of plugging in and checking, you'll find that y = 1 when x = 0... And y tends to x as x tends to infinity. So the lower limit is x + y = 1. And upper limit is infinite!

So if you do that final integral with limits 1 to infinite, you will get 0.75. Phew!

The polynomial $y^3 + y^2 x - x^2 y - x^3$ factors to $(y - x)(y + x)^2$ . Let $u = y - x$ . The polynomial is then $u(u + 2x)^2$ , so $u(u + 2x)^2 = 1$ , or $u = (u + 2x)^{-2}$ . $\frac{d \sqrt{u}}{dx} = \frac{d}{dx} (u + 2x)^{-1} = -(u + 2x)^{-2}(\frac{du}{dx} + 2) = -u(\frac{du}{dx} + 2) = -u \frac{du}{dx} - 2u$ . Thus, $\sqrt{u} = -\int u \frac{du}{dx} dx - 2 \int udx$ , so $\int udx = \frac{1}{2}(\int udu - \sqrt{u}) = \frac{1}{2}(\frac{u^2}{2} - \sqrt{u})$ . Now, to evaluate $\int_0^{\infty} (y - x) dx$ , the limits just need to be changed accordingly when substituting in $u$ . The equation $u = (u + 2x)^{-2}$ shows that if $x = 0$ , then $u = 1$ , and as $x \rightarrow \infty$ , $u \rightarrow 0$ . Therefore, $\int_0^{\infty} (y - x) dx = \int_1^0 u dx = - \int_0^1 u dx = -\frac{1}{2}(\frac{1^2}{2} - \sqrt{1}) - \frac{1}{2}(\frac{0^2}{2} - \sqrt{0}) = \frac{1}{2}(1 - \frac{1}{2}) = \frac{1}{2} \cdot \frac{3}{2} = \mathbf{\frac{3}{4}}$ .