Integrate ln(x^2) please, thanks!

Please show all the working as I'm not really sure about the integration of the natural logarithm, other than that integration ln (x) = x ln(x) - x ( + c). Thanks!

Use the formula you already have:

ln (x^2) = 2 ln x

∫ ln (x^2) dx = 2 ∫ ln x dx = 2 x ln x - 2x + C

Don't panic! Your integration of ln x was correct....and, in fact, it's an essential part of this problem.

Look carefully at ln (x^2), and think back to last year's math class. Can you think of a basic log property that can simplify this?

ln (x^2) = 2 ln(x)
Let us integrate ln(x) and multiply the result by 2

∫ ln(x) dx

Integrate by parts
dv = dx
v = x
u = ln (x)
du = 1/x dx

∫ u dv = uv - ∫ vdu
∫ ln(x) dx = x ln(x) - ∫ x (1/x) dx = x ln(x) - ∫ dx = x ln(x) - x

Therefore, ∫ 2 ln(x) dx = 2x ln(x) - 2x + C

You write Ln(x^2) = 2Ln(x) then
∫ ln (x) dx = x ln(x) - x + c
Or
∫ ln (x^2) dx = 2x ln(x) - 2x +C
May be you would wrote :
∫( ln (x) )^2 dx =
∫Ln^2(x) dx = xLn^2(x) - ∫x*1/x*2Ln(x)dx =
xLn^2(x) - 2∫Ln(x)dx =
xLn^2 (x) - 2x ln(x) + 2x +C
May be...

you cain't do this! don't ask here budddy , that's cheating. i'm sure ur parents anad teachers tot you better den dis.

imma hafta agree! thumbs up if you do and imma never stoP!