Integral of (ln(x))² geometrized.

We can calculate the red area by using our observation from above, replacing h with a differential, and integrating that orange triangle's area formula:

We can however, also subtract the integral from 0 to eᵐ of our blue curve equation, xeˣ, from the area inside the green outlined triangle to get our red area as well...

For the area of our green triangle, we can use the half-product of base and height, the base is the distance from the origin to the intercept of the vertical orange line, or eᵐ, and we can plug that into either our mx equation or xeˣ equation to get the height, which will be m ln(m). We can combine ln(m) · m ln(m) to m(ln(m))² to make it cleaner. We can then subtract the xeˣ integral to get the red area, which will be the same as what we calculated above:

...We can then integrate by parts:

***The limit of xeˣ as x approaches -∞ is 0 by domination of the exponential.

I will then consolidate this expression:

We used two different expressions to calculate the same red area, meaning the are equal:

We can the multiply both of these expressions by 2 to cancel the 1/2 coefficient, and just get the 'raw' integral of (ln(t))²: