Is there a different approach to evaluate $\int \ln(x)\,\mathrm{d}x?$

You could always eliminate the log by substituting $x=e^u$. We have $$\int\ln x\,dx=\int ue^u\,du\ .$$ Now you will still need integration by parts, but in this case it's a pretty obvious integration by parts which does not rely on the "trick" of inserting a factor of $1$.

By trial solution

$$F(x)=a(x)\log x+b(x) \implies F’(x)=a’(x) \log x+\frac{a(x)}{x}+b’(x)$$

then

• $a’(x)=1\implies a(x)=x$

• $b’(x)=-1\implies b(x)=-x$

Rough reasoning: \begin{align*} \ln x&=\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{1}{n}(x-1)^{n}\\ \int\ln xdx&=\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{1}{n}\cdot\dfrac{1}{n+1}(x-1)^{n+1}\\ &=\sum_{n=1}^{\infty}(-1)^{n+1}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)(x-1)^{n+1}\\ &=(x-1)\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{1}{n}(x-1)^{n}+\sum_{n=1}^{\infty}(-1)^{n+2}\dfrac{1}{n+1}(x-1)^{n+1}\\ &=(x-1)\ln x+\ln x-(x-1)\\ &=x\ln x-x+c. \end{align*}

Since for $t>1$ (the other case is similar) $\int_1^t \ln x dx$ is the (positive) area of the region $D\subset \mathbb R^2$ enclosed by the curve $y=\ln x$, the straight line $x=t$ and the horizontal axis, this area can also be written as a double integral $\iint_D dA$, which by Fubini theorem equals the iterated integral $$\int_1^t \int_0^{\ln x} dy\, dx.$$

Changing order of integration by writing $D=\{(x,y)\in\mathbb R^2 \colon 0\le y\le \ln t \wedge e^y \le x \le t\}$, that integral is equal to $$\int_0^{\ln t} \int_{e^y}^t dx\,dy=\int_0^{\ln t} (t-e^y) \;dy=(ty-e^y)|_{y=0}^{\ln t}=t \ln t -t+1.$$

That is, $$\int \ln x \; dx= x \ln x -x+C.$$