Let functions of x, s and t, be s=x^3, and t=log x (log: natural logarithm),so s'=3x^2 and t'=1/x.

Integrate the function given in product form, x^3·1/x=s·t', by parts formula shown below:

∫s·t'·dx = s·t - ∫s'·t·dx + C (The constant of integration C is skipped in the process.)

Substitute s,t,s' and t' in the equation, integration by parts formula, shown above.

We have: ∫x^3·1/x·dx = x^3·log x - 3∫x^2·log x·dx While, ∫x^3·1/x·dx = ∫x^2·dx = x^3/3

Thus. x^3/3 = x^3·log x - 3∫x^2·log x·dx

We have: ∫x^2·log x·dx=x^3/3·(log x - 1/3) + C

From this result, we assume that the following equqtion would hold valid as an identity:

∫x^n·log x·dx = 1/(n+1)·x^(n+1)·(log x - 1/(n+1)) + C, n ≠ -1

Plug n=3 into the equation above.

We have: ∫x^3·log x·dx = x^4/4·(log x - 1/4) + C

CK: Let t'=x^3 and s=log x, so t=∫x^3·dx=x^4/4, t'=x^3=(x^4/4)', and s'=1/x

Thus, ∫x^3·log x·dx = ∫(x^4/4)'·log x·dx = x^4/4·log x - ∫x^4/4·1/x·dx

= x^4/4·log x - 1/4·∫x^3·dx = x^4/4·log x - 1/4·x^4/4 = x^4/4·(log x - 1/4)

We have: ∫x^3·log x·dx = x^4/4·(log x - 1/4) + C CKD.

The answer is: ∫x^3·log x·dx = x^4/4·(log x - 1/4) + C