<s>[INST] Integral $\int\frac{dx}{x\sqrt{x^2+4}}$ While solving this:
$$
\begin{align}
\int\frac{dx}{x\sqrt{x^2+4}}
&=\int\frac{t(-1/t^2)dt}{\sqrt{(1/t)^2+4}}\tag{t=1/x}\\
&=\int\frac{(-1/t)dt}{(1/t)\sqrt{1+4t^2}}\\
&=-\frac12\int\frac{dt}{\sqrt{t^2+1/4}}\\
&=-\frac12\ln(t+\sqrt{t^2+1/4})+C\\
&=-\frac12\ln(1/x+\sqrt{1/x^2+1/4})+C\\
&=-\frac12\ln\left(\frac{2+\sqrt{4+x^2}}{2x}\right)+C\\
&=\color{green}{\frac12(\ln(x)-\ln(2+\sqrt{4+x^2}))}+C\tag{$\ln(2x)=\ln(x)+\ln2$}\\
\end{align}$$
I got this, but answer given in textbook is $\displaystyle\color{blue}{ \ln[(x+1/2)+\sqrt{x^2+x+1}]}+C$
I see no relation  between them, is my answer wrong? Along with answering this question please shed some light on this related issue which can happen with other integrals and etc.
 [/INST] You can see that your answer is correct and that the given answer is wrong because
$$\left(\frac 12\left(\ln (x)-\ln\left(2+\sqrt{4+x^2}\right)\right)\right)'=\frac{1}{x\sqrt{x^2+4}}$$
$$\left(\ln\left[(x+1/2)+\sqrt{x^2+x+1}\right]\right)'=\frac{1}{\sqrt{x^2+x+1}}$$
</s>