I need the antiderivative of this problem..?

Question

[(ln x)^2]/x

*that'd be (ln x) squared divided by x. Thank you so much in advance... i'm so freakin' lost on this one!!!

Answer 1

Use u substitution
u = ln x
du = 1/x dx

Plugging in,
integral u^2 du

= u^3/3
= (ln x)^3/3 + c

Answer 2

Try a U substitution.

If U = ln(x)
Then dU/dx = 1/x
and [(ln x)^2]/x·dx = U^2·du
The antiderivative of U^2 is U^3/3 + C
so the antiderivative is ln (x)^3/3 + C

(ln (x)^3/3 +C)' = 1/3·3ln(x)^2·1/x = ln(x)^2/x

Answer 3

I = ∫ [(ln x)² / x ] dx
let u = ln x
du = (1/x).dx
I = ∫ u² du
I = u³ / 3 + C
I = (ln x)³ / 3 + C