Please don't answer with other integration strategies, I'm trying to understand my error. Thanks.
I'm trying to integrate this
$$\int\frac{dx}{x\ln x\sqrt{\ln^2x+2\ln x + 4}}.$$
My try: letting $\ln x=y$, we have that $\frac{dx}{x}=dy$ so
$$\int\frac{dx}{x\ln x\sqrt{\ln^2x+2\ln x + 4}}=\int\frac{dy}{y\sqrt{y^2+2y+4}}=\int\frac{dy}{y\sqrt{\left(y+1\right)^2+3}}.$$
Now let $y+1=\sqrt3\sinh z,$ so $dy=\sqrt3 \cosh z \ dz$:
$$\int\frac{dy}{y\sqrt{\left(y+1\right)^2+3}}=\int\frac{\sqrt3 \cosh z}{(\sqrt3\sinh z-1)\sqrt{3\sinh^2 z +3}}dz=\int\frac{dz}{\sqrt3\sinh z -1}=$$
$$=\int\frac{dz}{\sqrt3\left(\frac{e^z-e^{-z}}{2}\right)-1}=2\int\frac{dz}{\sqrt3(e^z-e^{-z})-2}=2\int\frac{e^z}{\sqrt3e^{2z}-2e^z-\sqrt3}dz.$$
Let $e^z=t,$ so $e^z\ dz=dt$:
$$2\int\frac{e^z}{\sqrt3e^{2z}-2e^z-\sqrt3}dz=2\int\frac{dt}{\sqrt3t^2-2t-\sqrt3}=2\int\frac{dt}{\left(t-\sqrt3\right)\left(t+\frac{1}{\sqrt3}\right)}=$$
$$=\frac{2}{\sqrt3+\frac{1}{\sqrt3}}\int\left(\frac{1}{t-\sqrt3}-\frac{1}{t+\frac{1}{\sqrt3}}\right)dt=\frac{\sqrt3}{2}\ln \left|\frac{t-\sqrt3}{t+\frac{1}{\sqrt3}}\right|+c.$$
Recalling the substitution we have that
$$\frac{\sqrt3}{2}\ln \left|\frac{t-\sqrt3}{t+\frac{1}{\sqrt3}}\right|+c=\frac{\sqrt3}{2}\ln \left|\frac{e^{\sinh^{-1}\left(\frac{\ln x+1}{\sqrt3}\right)}-\sqrt3}{e^{\sinh^{-1}\left(\frac{\ln x+1}{\sqrt3}\right)}+\frac{1}{\sqrt3}}\right|+c$$
for all $c\in \mathbb{R}$, but according to integral calculator this isn't a correct antiderivative. Can someone spot the error please? I can't find it out.
You are missing a factor in this step: $$2\int\frac{dt}{\sqrt3t^2-2t-\sqrt3}=2\int\frac{dt}{\left(t-\sqrt3\right)\left(t+\frac{1}{\sqrt3}\right)}.$$ Expanding in the denominator yields $$t^2 + \left(\frac{1}{\sqrt{3}} - \sqrt{3}\right)t - 1.$$ This was the first error I found, after which I did not continue to check.