$\int (\ln x)^n dx$
in the most simplified form. Take care to compact long sums into $\sum$
expressions.

 Find
$\int xe^x dx.$
 Let
$a_n=n^2+3^na_{n1},$
with$a_0=1.$
Give a closed form expression for$a_n$
using sigma notation.
 Find

How could you find
$\int (\ln x)\ dx?$

How about treating the integrand as
$1\cdot \ln x?$

… and using integration by parts.

Why not apply this same technique to the original integral?

Try treating it as a recurrence relation.

… and unravelling it to obtain a
$\sum$
expression.

Let
$I_n=\int (\ln x)^ndx.$
We can obtain a recursive expression using integration by parts. We treat the integrand as the product$1 \cdot (\ln x)^n$
which gives us$I_n= x \ln^n(x)  \int n\ln^{n1}(x)\ dx,$
and we get the recurrence relation$I_n = x \ln^n(x)  n I_{n1}.$
We can expand out some of the terms to establish the pattern:$$ I_n= x \ln^n(x)  xn\ln^{n1}(x) + xn(n1)\ln^{n2}(x)\cdots $$
With each term, the sign flips, the exponent on the logarithm decreases by
$1,$
and each term collects factors decreasing from$n.$
Expressing this as a$\Sigma$
expression, we get$I_n=\sum_{i=0}^n (1)^i\frac{n!}{(ni)!}x\ln^{ni}(x).$