∫(x-1∕x)² dx Help, please

The question is in the title.
So far I have written it as
∫x²-(x^-1)²dx
and I am struggling with integrating the x^-1 as I don't know if it would be 2lnx or something else.
I've done the first x
∫x² dx =x³/3

The question is in the title.
So far I have written it as
∫x²-(x^-1)²dx
and I am struggling with integrating the x^-1 as I don't know if it would be 2lnx or something else.
I've done the first x
∫x² dx =x³/3

Expand $(x- \frac{1}{x})^2 = (x-\frac{1}{x})(x-\frac{1}{x})$

The result of $(a+b)^2$ is $a^2 +2ab +b^2$, that is the first term squared, plus two times the product, plus the last term squared.

$\int (x- \frac{1}{x})^2 dx = \int(x^2 - 2 + \frac{1}{x^2})dx$

From here it can be integrated easily.

Any help in what I did wrong?

Yeah, $(x-\frac{1}{x})^2 \not= x^2-(x^{-1})^2$

You can't go from ∫(x-1∕x)² to ∫x^2 - (x^-1)^2. You need to use the reverse chain rule so you first need to differentiate the bracket so you get dy/dx = 1 + x^-2. Using the reverse chain rule you get ((x-1/x)^3)/(3(1+x^-2)). Any math geniuses agree? Quite new to integration so sorry for any mistake. Hope it helps

Any help in what I did wrong?

x is a variable factor, so the reverse chain rule doesn't apply. You can divide by constants when using the reverse chain rule, but not by variable factors.

Expand $(x- \frac{1}{x})^2 = (x-\frac{1}{x})(x-\frac{1}{x})$

The result of $(a+b)^2$ is $a^2 +2ab +b^2$, that is the first term squared, plus two times the product, plus the last term squared.

$\int (x- \frac{1}{x})^2 dx = \int(x^2 - 2 + \frac{1}{x^2})dx$

From here it can be integrated easily.

Could you use the reverse chain rule, how would you use it in this question?