How can I integrate ln(x)^2 / x? The answer is 1/3 (lnx)^3 + C and I don't understand why.?

Question

I have a pretty good grasp of integration but don't get this question! Thanks!

Answer 1

To perhaps elaborate on nardneran's answer:

In this problem you use the substitution rule. Let u=ln x. Then du=1/x dx and dx=x du. Replacing ln x with u and dx with x du gives ∫u² du (the x's in your equation cancel). You can integrate that to get u³/3 + C and back-substituting gives (ln x)³/3 + C.

Answer 2

Let u=ln(x). Then du=(1/x) dx, so you get the integral of u^2 du.

Answer 3

integral of ln(x)^2/x is very very simple...

its intergral ln(x)^2/x dx [ and 1/x dx can be written as d( ln(x) ) ]

so its of the form f(x) d( f(x) )

so reduces to Y^2 dY where Y = ln(x)

there fore the answer is ln(x)^3/3 + C

Answer 4

considering the ques. take log x =t
differentiate both sides
1/x= dt/dx

y= t^2 dt/dx

now integrate
y= 1/3 t^3 + c

y= 1/3 (logx)^3 +c

Answer 5

If you differentiate your answer, you will see why. The +C is obviously required as a constant of integration as you do not have limits for x expressed in the question.

Answer 6

AHHHHHHHHHHH..Damn..Now my brain hurts... I knew I shouldnt have clicked on the mathmatics section!

Answer 7

its the quotient rule...