I don't speak Romanian, but the closest translation for this suggests you're trying to compute

Integrate by parts:

where
u = ln(x)² ⇒ du = 2 ln(x)/x dx
dv = x³ dx ⇒ v = 1/4 x⁴

Integrate by parts again:

where
u' = ln(x) ⇒ du' = dx/x
dv' = x³ dx ⇒ v' = 1/4 x⁴

So, we have





Answer:
Associative property is illustrated.
Step-by-step explanation:
we have been given:
(2+3.4)+6=2+(3.4+6)
This is the associative identity which is:
a+(b+c)=(a+b)+c
Here, we have a=2, b=3.4, c=6
On comparing the values with associative property we get:
(2+3.4)+6=2+(3.4+6)
We club two numbers b and c first and then a and b in same bracket.