$$ \int\left[\frac{\log x-1}{1+(\log x)^2}\right]\mathrm{d}x $$
Set $t=\log x\implies\mathrm{d}t=\dfrac{\mathrm{d}x}x$ $$ \int\frac{\log x-1}{1+(\log x)^2}\,\mathrm{d}x=\int\frac{t-1}{t^2+1}\mathrm{e}^t\,\mathrm{d}t\\ \frac{\mathrm{d}}{\mathrm{d}t}\frac1{t^2+1}=\frac{-2t}{(t^2+1)^2} $$
I think I am stuck, so is it the right substitution to obtain the solution or is there a better way to solve the given integral ?
My reference gives the solution $\dfrac{x}{1+(\log x)^2}$
Note: My reference is not that reliable, possible typo can be expected either in the question or in the solution.
Let's follow @Ak19's suggestion that we're trying to integrate $\frac{(t-1)^2}{(t^2+1)^2}e^t$ instead. A sensible Ansatz is $\frac{f(t)}{t^2+1}e^t$. So$$\frac{(t-1)^2}{(t^2+1)^2}e^t=\frac{d}{dt}\frac{f(t)e^t}{t^2+1}=\frac{f(t)e^t}{t^2+1}\left(\frac{f^\prime(t)}{f(t)}-\frac{2t}{t^2+1}+1\right),$$which simplifies to$$f^\prime(t)+\frac{(t-1)^2}{t^2+1}f(t)=\frac{(t-1)^2}{t^2+1}.$$Obviously, $f=1$ is a solution, which recovers the sought antiderivative. For more general ones, we of course add $+C$.
If this argument smacks of one we'd only make if we already know the answer, note that an Ansatz $g(t)e^t$ would work if$$g^\prime+g=\frac{(t-1)^2}{(t^2+1)^2}=\frac{1}{t^2+1}-\frac{2t}{(t^2+1)^2},$$so we can take $g=\frac{1}{t^2+1}$.