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*}