reduction formula

(i)
I_n
=∫x^3(lnx)^n dx
=(1/4)∫(lnx)^n d(x^4)
=(1/4)(lnx)^n(x^4) - (1/4)∫n(x^3)(lnx)^(n-1)dx
=(1/4)(ln e)^n(e^4) - (n/4)I_n-1
=e^4/4 - (n/4)I_n-1
(ii)
I_0=∫x^3dx=x^4/4=(e^4-1)/4
The original integration is I_2
I_2
=e^4/4 - (2/4)I_1
=e^4/4 - (1/2)(e^4/4 - (1/4)I_0)
=e^4/4 - (1/2)(e^4/4 - (1/16)(e^4-1))
=e^4/4-e^4/8+e^4/32-1/32
=(5e^4-1)/32