Integrate $\ln (x)$ without integration by parts I know with integration by parts the answer is $x\ln(x) - x$ but I was wondering how to do this without integration by parts.

Compute it via the Riemann sums over the subdivision $x_k=aq^k$ with $x_N=aq^N=b$. Then
\begin{align}
\int_a^b\ln(x)dx=\lim_{N\to\infty}\sum_{n=0}^{N-1}(\ln(a)+n\ln(q))aq^n(q-1)
\end{align}
and using the sum formulas for geometric sums $\sum_{n=0}^{N-1}q^n=\frac{q^N-1}{q-1}$ and $\sum_{n=0}^{N-1}nq^n=\frac{q-Nq^N+(N-1)q^{N+1}}{(q-1)^2}$  we get
\begin{align}
\sum_{n=0}^{N-1}(\ln(a)+n\ln(q))aq^n(q-1)
&=a(q^N-1)\ln(a)+a\frac{q-Nq^N+(N-1)q^{N+1}}{q-1}\ln(q)  \\
&=(b-a)\ln(a)+\frac{qa-Nb+(N-1)qb}{q-1}\ln(q)  \\
&=b(\ln(a)+N\ln(q))-a\ln(a)+\frac{q(a-b)}{q-1}\ln(q)  \\
&=b\ln(b)-a\ln(a)-q(b-a)\frac{\ln(q)-\ln(1)}{q-1}
\end{align}
Now for $N\to\infty$ we get $q\to 1$ and thus $\ln(q)\to0$, and in the last term the difference quotient converges to the derivative $1$ in $q=1$, so that
$$
\int_a^b\ln(x)\,dx=b\ln(b)-a\ln(a)-(b-a)
$$
remains.
