High school integration problem I love doing mathématique which are a little bit hard for my level to challenge myselft and push my mathématiques boundaries. And I come across a problem I can't understand how to finish.
I have to prove that : 
$$\left |f(x) - \frac{1}{2x}  \right | = \int_{x}^{2x} \frac{t^2 + 1}{t^2\sqrt{t^4+t^2+1}(t^2 + \sqrt{t^4+t^2+1})}dt$$
with, 
$$f(x) = \int_{x}^{2x} \frac{1}{\sqrt{t^4+t^2+1}}dt$$


The answer is : 
$$\left |f(x) - \frac{1}{2x}  \right | = \left | \int_{x}^{2x} \frac{1}{\sqrt{t^4+t^2+1}}dt - \int_{x}^{2x} \frac{1}{t^2}dt \right |$$
To obtaint the final form, you have to develop and do some integration calculus, not very hard.

I don't understand how to find that :
$$\frac{1}{2x} = \int_{x}^{2x} \frac{1}{t^2}dt$$
 A: Write $\dfrac1{t^2}$ as $t^{-2}$ and apply the power rule for integrals.
$$\int_x^{2x} \dfrac1{t^2} \, dt = \int_x^{2x} t^{-2} \, dt = -t^{-1} \bigg|_x^{2x} = -\frac1t\bigg|_x^{2x} = -\left(\frac1{2x}- \frac1x\right) = \frac1{2x} $$
A: This is simply the integration of a polynomial $t^{-2}$, so just use the rule of "add 1 to the power, divide by the new power" to get the integral to be $-t^{-1}$.
$$\int\frac1{t^2}\,dt=-\frac1t+c$$
Applying limits of $2x$, $x$: $$\left[c-\frac1t\right]_x^{2x}=\left(c-\frac1{2x}\right)-\left(c-\frac1x\right)=\frac1x-\frac1{2x}=\frac{2-1}{2x}=\frac1{2x}$$

EDIT:
Oh I see, so you want to jump from having $\frac{1}{2x}$ back to an integral? This would probably need a bit of guess work - you see that $f(x)$ has an integral with limits $2x$ and $x$, so you can set this up:
$$\frac{1}{2x}=\int_x^{2x}g(t)\,dt$$ for some $g(t)$. Then suppose $g(t)$ has antiderivative $G(t)$, so we get $$\frac1{2x}=\int_x^{2x}G'(t)\,dt=\left[G(t)\right]_x^{2x}=G(2x)-G(x)$$
Then solving this (I am not sure how to do so systematically) would give you the solution you're after. I'd say to try a polynomial solution since $\frac1{2x}$ is a power of $x$. So try $G(x)=Ax^\alpha\implies 2^\alpha Ax^\alpha-Ax^\alpha=\frac1{2x}\implies \alpha=-1, \frac A2-A=\frac12\implies A=-1$.
Thus $G(t)=-\frac1t\implies g(t)=\frac 1{t^2}$ works.
