How do I find this integral? x^3*ln xdx?

Answer 1

Derek C appears to have it right.

Check for yourself on the following mathematics page:
http://integrals.wolfram.com/index.jsp

Answer 2

Use integration by parts.
Let u = ln x and v' = x^3, so that u' = 1/x and v = x^4/4.

The integral of uv' = uv - integral of vu'

so the integral becomes:

x^4/4 * ln(x) - integral of (x^4/4)(1/x) dx
= x^4/4 * ln x - integral of x^3 / 4 dx
= x^4/4 * ln x - x^4/16 + C
= (1/16)x^4 (4 ln x - 1) + C.

Answer 3

Put u = ln(x), du = dx/x and dv = x^3 dx. Then, by parts,

Int x^3*ln(x) dx = (x^4/4) ln(x) - Int (x^4/4) * (1/x) dx = (x^4/4) ln(x) - 1/4 Int x^3 dx = (x^4/4) ln(x) - x^4/16 + C, = (x^4/4) [ln(x) -1/4] C the integration constant.

Answer 4

Integration by parts.

Let u = lnx and dv = x^3 dx
du = 1/x dx and v = (x^4)/4

∫ [(lnx) • x^3] dx = ∫ u dv = uv - ∫ (v du) dx + C =

(lnx)(x^4)/4 - ∫ (1/x)•(x^4)/4 dx + C =
(lnx)(x^4)/4 - ∫ (x^3)/4 dx + C =

(lnx)•(x^4)/4 - (x^4)/16 + C