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.
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You can see that your answer is correct and that the given answer is wrong because $$\left(\frac 12\left(\ln (x)-\ln\left(2+\sqrt{4+x^2}\right)\right)\right)'=\frac{1}{x\sqrt{x^2+4}}$$ $$\left(\ln\left[(x+1/2)+\sqrt{x^2+x+1}\right]\right)'=\frac{1}{\sqrt{x^2+x+1}}$$
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