Evaluate the definite integral of [(ln x)^2 dx]/x with lower limit 0 and upper limit 3?

Answer 1

[(ln x)^2 dx]/x

Let u = ln(x)
du/dx = 1/x
du = (1/x)dx

[(ln x)^2 dx]/x
= u^2 du

Integrate this to get (1/3)u^3, which is equivalent to (1/3)[ln(x)]^3

(1/3)(ln x)^3

Apply the limits.
Upper limit: 3 ---- (1/3)(ln 3)^3
Lower limit: 0 ---- ln(0) approaches -infinity

(1/3)(ln 3)^3 - (-infinity) = infinity

The integral diverges.

Answer 2

that is impossible due to the lower limit being zero.
integral of [(ln x)^2 dx]/x is [(ln x)^3]/3 using integration by parts. putting the limits we have [(ln 3)^3 - (ln 0)^3]/3. but ln 0 doesn't exist since ln x is defined for x>0

Answer 3

The former answer states the indefinite integral of the function
but the definite integral(0,3) does not exist as lim
of the indefinite Integral x==>0 is infinite