Can anyone tell me the steps and answer to the integral:x^2ln(3x)dx?

Answer 1

You can use the integration by parts:

Integral[x^2*Log[3x]] =
= 1/3*x^3*Log[3x] - Integral[1/3*x^3*1/(3x)] =
= 1/3*x^3*Log[3x] - Integral[1/3*x^2] =
= 1/3*x^3*Log[3x] - 1/9*x^3 =
= 1/9*x^3(3*Log[3x]-1)

Answer 2

Substitute:

z = 3x ; x² = z² / 9; dx = dz

∫x² ln |3x| dx = (1/9) ∫ z² ln |z| dz

integrate by parts:

u = ln |z|; dv = z² dz

du = dz / z; v = z³ / 3
-----------------------------------------
(1/9) ∫ z² ln |z| dz = (1/27) z³ ln |z| - (1/27) ∫ z² dz

= (1/27) z³ ( ln |z| - 1/3)

Back substitute:

(1/27) z³ ( ln |z| - 1/3) = x³ ( ln |3x| - 1/3)

Makes sense?
~W.O.M.B.A.T.

Answer 3

integrate by parts
u=ln(3x), du = 1/x dx
dv = x^2 dx, v = x^3/3

so the integral is uv - int(vdu)
= (x^3 * ln(3x))/3 - int(x^2/3 dx)
= (x^3 * ln(3x))/3 - x^3/9
and don't forget the "+C".

Answer 4

now do it ur self